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12.2: Vectors in Three Dimensions - Mathematics

Learning Objectives

• Describe three-dimensional space mathematically.
• Locate points in space using coordinates.
• Write the distance formula in three dimensions.
• Write the equations for simple planes and spheres.
• Perform vector operations in (mathbb{R}^{3}).

Vectors are useful tools for solving two-dimensional problems. Life, however, happens in three dimensions. To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases the topography of the land is important. Does your planned route go through the mountains? Do you have to cross a river? To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a natural extension of the two-dimensional Cartesian coordinate plane into three dimensions.

Three-Dimensional Coordinate Systems

As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal (x)-axis and the vertical (y)-axis. We can add a third dimension, the (z)-axis, which is perpendicular to both the (x)-axis and the (y)-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions we encounter in real life.

Definition: Three-dimensional Rectangular Coordinate System

The three-dimensional rectangular coordinate system consists of three perpendicular axes: the (x)-axis, the (y)-axis, and the (z)-axis. Because each axis is a number line representing all real numbers in (ℝ), the three-dimensional system is often denoted by (ℝ^3).

In Figure (PageIndex{1a}), the positive (z)-axis is shown above the plane containing the (x)- and (y)-axes. The positive (x)-axis appears to the left and the positive (y)-axis is to the right. A natural question to ask is: How was this arrangement determined? The system displayed follows the right-hand rule. If we take our right hand and align the fingers with the positive (x)-axis, then curl the fingers so they point in the direction of the positive (y)-axis, our thumb points in the direction of the positive (z)-axis (Figure (PageIndex{1b})). In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the standard representation.

In two dimensions, we describe a point in the plane with the coordinates ((x,y)). Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate, (z), is appended to indicate alignment with the (z)-axis: ((x,y,z)). A point in space is identified by all three coordinates (Figure (PageIndex{2})). To plot the point ((x,y,z)), go (x) units along the (x)-axis, then (y) units in the direction of the (y)-axis, then (z) units in the direction of the (z)-axis.

Example (PageIndex{1}): Locating Points in Space

Sketch the point ((1,−2,3)) in three-dimensional space.

Solution

To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive (x) direction, (2) units in the negative (y) direction, and (3) units in the positive (z) direction. Complete the prism to plot the point (Figure).

Exercise (PageIndex{1})

Sketch the point ((−2,3,−1)) in three-dimensional space.

Hint

Start by sketching the coordinate axes. e.g., Figure (PageIndex{3}). Then sketch a rectangular prism to help find the point in space.

In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a coordinate plane: the (xy)-plane, the (xz)-plane, and the (yz)-plane (Figure (PageIndex{3})). We define the (xy)-plane formally as the following set: ({(x,y,0):x,y∈ℝ}.) Similarly, the (xz)-plane and the (yz)-plane are defined as ({(x,0,z):x,z∈ℝ}) and ({(0,y,z):y,z∈ℝ},) respectively.

To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the (xy)-plane, the wall to your right is the (xz)-plane, and the wall to your left is the (yz)-plane.

In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants. The octants fill (ℝ^3) in the same way that quadrants fill (ℝ^2), as shown in Figure (PageIndex{4}).

Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space.

If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We that the distance (d) between two points ((x_1,y_1)) and ((x_2,y_2)) in the x(y)-coordinate plane is given by the formula

[d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2}.]

The formula for the distance between two points in space is a natural extension of this formula.

The Distance between Two Points in Space

The distance (d) between points ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)) is given by the formula

[d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}. label{distanceForm}]

The proof of this theorem is left as an exercise. (Hint: First find the distance (d_1) between the points ((x_1,y_1,z_1)) and ((x_2,y_2,z_1)) as shown in Figure (PageIndex{5}).)

Example (PageIndex{2}): Distance in Space

Find the distance between points (P_1=(3,−1,5)) and (P_2=(2,1,−1).)

Solution

Substitute values directly into the distance formula (Equation ef{distanceForm}):

[egin{align*} d(P_1,P_2) &=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2} [4pt] &=sqrt{(2−3)^2+(1−(−1))^2+(−1−5)^2} [4pt] &=sqrt{(-1)^2+2^2+(−6)^2} [4pt] &=sqrt{41}. end{align*}]

Exercise (PageIndex{2})

Find the distance between points (P_1=(1,−5,4)) and (P_2=(4,−1,−1)).

Hint

(d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2})

(5sqrt{2})

Before moving on to the next section, let’s get a feel for how (ℝ^3) differs from (ℝ^2). For example, in (ℝ^2), lines that are not parallel must always intersect. This is not the case in (ℝ^3). For example, consider the line shown in Figure (PageIndex{7}). These two lines are not parallel, nor do they intersect.

Figure (PageIndex{7}): These two lines are not parallel, but still do not intersect.

You can also have circles that are interconnected but have no points in common, as in Figure (PageIndex{8}).

Figure (PageIndex{8}): These circles are interconnected, but have no points in common.

We have a lot more flexibility working in three dimensions than we do if we stuck with only two dimensions.

Writing Equations in (ℝ^3)

Now that we can represent points in space and find the distance between them, we can learn how to write equations of geometric objects such as lines, planes, and curved surfaces in (ℝ^3). First, we start with a simple equation. Compare the graphs of the equation (x=0) in (ℝ), (ℝ^2),and (ℝ^3) (Figure (PageIndex{9})). From these graphs, we can see the same equation can describe a point, a line, or a plane.

In space, the equation (x=0) describes all points ((0,y,z)). This equation defines the (yz)-plane. Similarly, the (xy)-plane contains all points of the form ((x,y,0)). The equation (z=0) defines the (xy)-plane and the equation (y=0) describes the (xz)-plane (Figure (PageIndex{10})).

Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one of the coordinate planes. When a plane is parallel to the (xy)-plane, for example, the (z)-coordinate of each point in the plane has the same constant value. Only the (x)- and (y)-coordinates of points in that plane vary from point to point.

Equations of Planes Parallel to Coordinate Planes

1. The plane in space that is parallel to the (xy)-plane and contains point ((a,b,c)) can be represented by the equation (z=c).
2. The plane in space that is parallel to the (xz)-plane and contains point ((a,b,c)) can be represented by the equation (y=b).
3. The plane in space that is parallel to the (yz)-plane and contains point ((a,b,c)) can be represented by the equation (x=a).

Example (PageIndex{3}): Writing Equations of Planes Parallel to Coordinate Planes

1. Write an equation of the plane passing through point ((3,11,7)) that is parallel to the (yz)-plane.
2. Find an equation of the plane passing through points ((6,−2,9), (0,−2,4),) and ((1,−2,−3).)

Solution

1. When a plane is parallel to the (yz)-plane, only the (y)- and (z)-coordinates may vary. The (x)-coordinate has the same constant value for all points in this plane, so this plane can be represented by the equation (x=3).
2. Each of the points ((6,−2,9), (0,−2,4),) and ((1,−2,−3)) has the same (y)-coordinate. This plane can be represented by the equation (y=−2).

Exercise (PageIndex{3})

Write an equation of the plane passing through point ((1,−6,−4)) that is parallel to the (xy)-plane.

Hint

If a plane is parallel to the (xy)-plane, the z-coordinates of the points in that plane do not vary.

(z=−4)

As we have seen, in (ℝ^2) the equation (x=5) describes the vertical line passing through point ((5,0)). This line is parallel to the (y)-axis. In a natural extension, the equation (x=5) in (ℝ^3) describes the plane passing through point ((5,0,0)), which is parallel to the (yz)-plane. Another natural extension of a familiar equation is found in the equation of a sphere.

Definition: Sphere

A sphere is the set of all points in space equidistant from a fixed point, the center of the sphere (Figure (PageIndex{11})), just as the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, the distance from the center to a point on the sphere is called the radius.

The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance.

Standard Equation of a Sphere

The sphere with center ((a,b,c)) and radius (r) can be represented by the equation

[(x−a)^2+(y−b)^2+(z−c)^2=r^2.]

This equation is known as the standard equation of a sphere.

Example (PageIndex{4}): Finding an Equation of a Sphere

Find the standard equation of the sphere with center ((10,7,4)) and point ((−1,3,−2)), as shown in Figure (PageIndex{12}).

Figure (PageIndex{12}): The sphere centered at ((10,7,4)) containing point ((−1,3,−2).)

Solution

Use the distance formula to find the radius (r) of the sphere:

[egin{align*} r &=sqrt{(−1−10)^2+(3−7)^2+(−2−4)^2} [4pt] &=sqrt{(−11)^2+(−4)^2+(−6)^2} [4pt] &=sqrt{173} end{align*} ]

The standard equation of the sphere is

[(x−10)^2+(y−7)^2+(z−4)^2=173. onumber]

Exercise (PageIndex{4})

Find the standard equation of the sphere with center ((−2,4,−5)) containing point ((4,4,−1).)

Hint

First use the distance formula to find the radius of the sphere.

[(x+2)^2+(y−4)^2+(z+5)^2=52 onumber]

Example (PageIndex{5}): Finding the Equation of a Sphere

Let (P=(−5,2,3)) and (Q=(3,4,−1)), and suppose line segment (overline{PQ}) forms the diameter of a sphere (Figure (PageIndex{13})). Find the equation of the sphere.

Solution:

Since (overline{PQ}) is a diameter of the sphere, we know the center of the sphere is the midpoint of (overline{PQ}).Then,

[C=left(dfrac{−5+3}{2},dfrac{2+4}{2},dfrac{3+(−1)}{2} ight)=(−1,3,1). onumber]

Furthermore, we know the radius of the sphere is half the length of the diameter. This gives

[egin{align*} r &=dfrac{1}{2}sqrt{(−5−3)^2+(2−4)^2+(3−(−1))^2} [4pt] &=dfrac{1}{2}sqrt{64+4+16} [4pt] &=sqrt{21} end{align*}]

Then, the equation of the sphere is ((x+1)^2+(y−3)^2+(z−1)^2=21.)

Exercise (PageIndex{5})

Find the equation of the sphere with diameter (overline{PQ}), where (P=(2,−1,−3)) and (Q=(−2,5,−1).)

Hint

Find the midpoint of the diameter first.

[x^2+(y−2)^2+(z+2)^2=14 onumber]

Example (PageIndex{6}): Graphing Other Equations in Three Dimensions

Describe the set of points that satisfies ((x−4)(z−2)=0,) and graph the set.

Solution

We must have either (x−4=0) or (z−2=0), so the set of points forms the two planes (x=4) and (z=2) (Figure (PageIndex{14})).

Exercise (PageIndex{6})

Describe the set of points that satisfies ((y+2)(z−3)=0,) and graph the set.

Hint

One of the factors must be zero.

The set of points forms the two planes (y=−2) and (z=3).

Example (PageIndex{7}): Graphing Other Equations in Three Dimensions

Describe the set of points in three-dimensional space that satisfies ((x−2)^2+(y−1)^2=4,) and graph the set.

Solution

The (x)- and (y)-coordinates form a circle in the (xy)-plane of radius (2), centered at ((2,1)). Since there is no restriction on the (z)-coordinate, the three-dimensional result is a circular cylinder of radius (2) centered on the line with (x=2) and (y=1). The cylinder extends indefinitely in the (z)-direction (Figure (PageIndex{15})).

Exercise (PageIndex{7})

Describe the set of points in three dimensional space that satisfies (x^2+(z−2)^2=16), and graph the surface.

Hint

Think about what happens if you plot this equation in two dimensions in the (xz)-plane.

A cylinder of radius 4 centered on the line with (x=0) and (z=2).

Working with Vectors in (ℝ^3)

Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.

Three-dimensional vectors can also be represented in component form. The notation (vecs{v}=⟨x,y,z⟩) is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, ((0,0,0)), and terminal point ((x,y,z)). The zero vector is (vecs{0}=⟨0,0,0⟩). So, for example, the three dimensional vector (vecs{v}=⟨2,4,1⟩) is represented by a directed line segment from point ((0,0,0)) to point ((2,4,1)) (Figure (PageIndex{16})).

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If (vecs{v}=⟨x_1,y_1,z_1⟩) and (vecs{w}=⟨x_2,y_2,z_2⟩) are vectors, and (k) is a scalar, then

[vecs{v}+vecs{w}=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩]

and

[kvecs{v}=⟨kx_1,ky_1,kz_1⟩.]

If (k=−1,) then (kvecs{v}=(−1)vecs{v}) is written as (−vecs{v}), and vector subtraction is defined by (vecs{v}−vecs{w}=vecs{v}+(−vecs{w})=vecs{v}+(−1)vecs{w}).

The standard unit vectors extend easily into three dimensions as well, (hat{mathbf i}=⟨1,0,0⟩), (hat{mathbf j}=⟨0,1,0⟩), and (hat{mathbf k}=⟨0,0,1⟩), and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in (ℝ^3) in the following ways:

[vecs{v}=⟨x,y,z⟩=xhat{mathbf i}+yhat{mathbf j}+zhat{mathbf k}].

Example (PageIndex{8}): Vector Representations

Let (vecd{PQ}) be the vector with initial point (P=(3,12,6)) and terminal point (Q=(−4,−3,2)) as shown in Figure (PageIndex{17}). Express (vecd{PQ}) in both component form and using standard unit vectors.

Solution

In component form,

[egin{align*} vecd{PQ} =⟨x_2−x_1,y_2−y_1,z_2−z_1⟩ [4pt] =⟨−4−3,−3−12,2−6⟩ [4pt] =⟨−7,−15,−4⟩. end{align*}]

In standard unit form,

[vecd{PQ}=−7hat{mathbf i}−15hat{mathbf j}−4hat{mathbf k}. onumber]

Exercise (PageIndex{8})

Let (S=(3,8,2)) and (T=(2,−1,3)). Express (vec{ST}) in component form and in standard unit form.

Hint

Write (vecd{ST}) in component form first. (T) is the terminal point of (vecd{ST}).

(vecd{ST}=⟨−1,−9,1⟩=−hat{mathbf i}−9hat{mathbf j}+hat{mathbf k})

As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space (Figure (PageIndex{18})).

We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.

Properties of Vectors in Space

Let (vecs{v}=⟨x_1,y_1,z_1⟩) and (vecs{w}=⟨x_2,y_2,z_2⟩) be vectors, and let (k) be a scalar.

• Scalar multiplication: [kvecs{v}=⟨kx_1,ky_1,kz_1⟩]
• Vector addition: [vecs{v}+vecs{w}=⟨x_1,y_1,z_1⟩+⟨x_2,y_2,z_2⟩=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩]
• Vector subtraction: [vecs{v}−vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩]
• Vector magnitude: [|vecs{v}|=sqrt{x_1^2+y_1^2+z_1^2}]
• Unit vector in the direction of (vecs{v}): [dfrac{1}{|vecs{v}|}vecs{v}=dfrac{1}{|vecs{v}|}⟨x_1,y_1,z_1⟩=⟨dfrac{x_1}{|vecs{v}|},dfrac{y_1}{|vecs{v}|},dfrac{z_1}{|vecs{v}|}⟩, quad ext{if} , vecs{v}≠vecs{0}]

We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

Example (PageIndex{9}): Vector Operations in Three Dimensions

Let (vecs{v}=⟨−2,9,5⟩) and (vecs{w}=⟨1,−1,0⟩) (Figure (PageIndex{19})). Find the following vectors.

1. (3vecs{v}−2vecs{w})
2. (5|vecs{w}|)
3. (|5 vecs{w}|)
4. A unit vector in the direction of (vecs{v})

Solution

a. First, use scalar multiplication of each vector, then subtract:

[egin{align*} 3vecs{v}−2vecs{w} =3⟨−2,9,5⟩−2⟨1,−1,0⟩ [4pt] =⟨−6,27,15⟩−⟨2,−2,0⟩ [4pt] =⟨−6−2,27−(−2),15−0⟩ [4pt] =⟨−8,29,15⟩. end{align*}]

b. Write the equation for the magnitude of the vector, then use scalar multiplication:

[5|vecs{w}|=5sqrt{1^2+(−1)^2+0^2}=5sqrt{2}. onumber]

c. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:

[|5 vecs{w}|=∥⟨5,−5,0⟩∥=sqrt{5^2+(−5)^2+0^2}=sqrt{50}=5sqrt{2} onumber]

d. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:

[egin{align*} dfrac{vecs{v}}{|vecs{v}|} =dfrac{1}{|vecs{v}|}⟨−2,9,5⟩ [4pt] =dfrac{1}{sqrt{(−2)^2+9^2+5^2}}⟨−2,9,5⟩ [4pt] =dfrac{1}{sqrt{110}}⟨−2,9,5⟩ [4pt] =⟨dfrac{−2}{sqrt{110}},dfrac{9}{sqrt{110}},dfrac{5}{sqrt{110}}⟩ . end{align*}]

Exercise (PageIndex{9}):

Let (vecs{v}=⟨−1,−1,1⟩) and (vecs{w}=⟨2,0,1⟩). Find a unit vector in the direction of (5vecs{v}+3vecs{w}.)

Hint

Start by writing (5vecs{v}+3vecs{w}) in component form.

(⟨dfrac{1}{3sqrt{10}},−dfrac{5}{3sqrt{10}},dfrac{8}{3sqrt{10}}⟩)

Example (PageIndex{10}): Throwing a Forward Pass

A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of (30°) (see the following figure). Write the initial velocity vector of the ball, (vecs{v}), in component form.

Solution

The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector (vecs{w}) extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of (30°) (see the following figure). This vector would have the same direction as (vecs{v}), but it may not have the right magnitude.

The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is

Dist from QB to receiver(=sqrt{15^2+20^2}=sqrt{225+400}=sqrt{625}=25) yd.

We have (dfrac{25}{|vecs{w}|}=cos 30°.) Then the magnitude of (vecs{w}) is given by

(|vecs{w}|=dfrac{25}{cos 30°}=dfrac{25⋅2}{sqrt{3}}=dfrac{50}{sqrt{3}}) yd

and the vertical distance from the receiver to the terminal point of (vecs{w}) is

Vert dist from receiver to terminal point of (vecs{w}=|vecs{w}| sin 30°=dfrac{50}{sqrt{3}}⋅dfrac{1}{2}=dfrac{25}{sqrt{3}}) yd.

Then (vecs{w}=⟨20,15,dfrac{25}{sqrt{3}}⟩), and has the same direction as (vecs{v}).

Recall, though, that we calculated the magnitude of (vecs{w}) to be (|vecs{w}|=dfrac{50}{sqrt{3}}), and (vecs{v}) has magnitude (60) mph. So, we need to multiply vector (vecs{w}) by an appropriate constant, (k). We want to find a value of (k) so that (∥kvecs{w}∥=60) mph. We have

(|k vecs{w}|=k|vecs{w}|=kdfrac{50}{sqrt{3}}) mph,

so we want

(kdfrac{50}{sqrt{3}}=60)

(k=dfrac{60sqrt{3}}{50})

(k=dfrac{6sqrt{3}}{5}).

Then

(vecs{v}=kvecs{w}=k⟨20,15,dfrac{25}{sqrt{3}}⟩=dfrac{6sqrt{3}}{5}⟨20,15,dfrac{25}{sqrt{3}}⟩=⟨24sqrt{3},18sqrt{3},30⟩).

Let’s double-check that (|vecs{v}|=60.) We have

(|vecs{v}|=sqrt{(24sqrt{3})^2+(18sqrt{3})^2+(30)^2}=sqrt{1728+972+900}=sqrt{3600}=60) mph.

So, we have found the correct components for (vecs{v}).

Exercise (PageIndex{10})

Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of (40) mph and an angle of (45°). Write the initial velocity vector of the ball, (vecs{v}), in component form.

Hint

Follow the process used in the previous example.

(v=⟨16sqrt{2},12sqrt{2},20sqrt{2}⟩)

Key Concepts

• The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples ((x,y,z)) are used to describe the location of a point in space.
• The distance (d) between points ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)) is given by the formula [d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}. onumber]
• In three dimensions, the equations (x=a,y=b,) and (z=c) describe planes that are parallel to the coordinate planes.
• The standard equation of a sphere with center ((a,b,c)) and radius (r) is [(x−a)^2+(y−b)^2+(z−c)^2=r^2. onumber ]
• In three dimensions, as in two, vectors are commonly expressed in component form, (v=⟨x,y,z⟩), or in terms of the standard unit vectors, (xi+yj+zk.)
• Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let (v=⟨x_1,y_1,z_1⟩) and (w=⟨x_2,y_2,z_2⟩) be vectors, and let (k) be a scalar.

Scalar multiplication:

[(kvecs{v}=⟨kx_1,ky_1,kz_1⟩ onumber]

[vecs{v}+vecs{w}=⟨x_1,y_1,z_1⟩+⟨x_2,y_2,z_2⟩=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩ onumber]

Vector subtraction:

[vecs{v}−vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩ onumber]

Vector magnitude:

[‖vecs{v}‖=sqrt{x_1^2+y_1^2+z_1^2} onumber]

Unit vector in the direction of (vecs{v}):

[dfrac{vecs{v}}{‖vecs{v}‖}=dfrac{1}{‖vecs{v}‖}⟨x_1,y_1,z_1⟩=⟨dfrac{x_1}{‖vecs{v}‖},dfrac{y_1}{‖vecs{v}‖},dfrac{z_1}{‖vecs{v}‖}⟩, vecs{v}≠vecs{0} onumber]

Key Equations

Distance between two points in space:

[d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}]

Sphere with center ((a,b,c)) and radius (r):

[(x−a)^2+(y−b)^2+(z−c)^2=r^2]

Glossary

coordinate plane
a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the (xy)-plane, (xz)-plane, or the (yz)-plane
right-hand rule
a common way to define the orientation of the three-dimensional coordinate system; when the right hand is curved around the (z)-axis in such a way that the fingers curl from the positive (x)-axis to the positive (y)-axis, the thumb points in the direction of the positive (z)-axis
octants
the eight regions of space created by the coordinate planes
sphere
the set of all points equidistant from a given point known as the center
standard equation of a sphere
((x−a)^2+(y−b)^2+(z−c)^2=r^2) describes a sphere with center ((a,b,c)) and radius (r)
three-dimensional rectangular coordinate system
a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple ((x,y,z)) that plots its location relative to the defining axes

Dimension (vector space)

In mathematics, the dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis of V over its base field. [1] [2] It is sometimes called Hamel dimension (after Georg Hamel) or algebraic dimension to distinguish it from other types of dimension.

The dimension of the vector space V over the field F can be written as dimF(V) or as [V : F], read "dimension of V over F". When F can be inferred from context, dim(V) is typically written.

Figures in Three Dimensions

We can still plot 2 dimensional shapes in 3 dimensions, all we need is to set any
of the axes to 0.

The formula for a circle in the xy plane would be

Notice that when z is 0, it is the same equation for a 2 dimensional circle. When
z does not equal 0, the graph will form a sphere.

The graph of a circle in the xy plane forms a cylinder when z can take on any value

Let’s graph x = 3 in one, two, and three dimensions. Here is x = 3 in R 1

Since x is the only dimension, the image is just a point on a number line.

Because y can take on any value when x is 3, the image is a straight line.

Here, y and z can take on any value, so the image is the yz plane at x =
3.

Notice that first we have a point, then a line, and then finally a plane.

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to three Dimensional Geometry Ex 12.2

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)
Solution:
(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is
(PQ=sqrt < left( 4-2 ight) ^< 2 >+left( 3-3 ight) ^< 2 >left( 1-5 ight) ^ < 2 >> )
= (sqrt < 4+0+16= >sqrt < 20 >=2sqrt < 5 >units.)

(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is
(PQ=sqrt < left[ 2-left( -3 ight) ight] ^< 2 >+left( 4-7 ight) ^< 2 >left( -1-2 ight) ^ < 2 >> )
(=sqrt < left( 2+3 ight) ^< 2 >+left( 4-7 ight) ^< 2 >+left( -1-2 ight) ^ < 2 >> )
(=sqrt < 25+9+9 >=sqrt < 43 >units)

(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is
(PQ=sqrt < left[ 1-left( -1 ight) ight] ^< 2 >+left( -3-3 ight) ^< 2 >left[ 4-left( -4 ight) ight] ^ < 2 >> )
(=sqrt < 4+36+64 >=sqrt < 104 >=2sqrt < 26 >units)

(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is
(PQ=sqrt < left( -2-2 ight) ^< 2 >+left[ 1-left( -1 ight) ight] ^< 2 >+left( 3-3 ight) ^ < 2 >> )
(=sqrt < 16+4+0 >=sqrt < 20 >=2sqrt < 5 >units)

Question 2.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.

Now AC = AB + BC
Thus, points A, B and C are collinear.

Question 3.
Verify the following:
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7,8) and (2, -3,4) are the vertices of a parallelogram.
Solution:
(i) Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then

Now, AB = BC
Thus, ABC is an isosceles triangle.

(ii) Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then

Now, AC 2 = AB 2 + BC 2
Thus, ABC is a right angled triangle.

(iii) Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then

Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.

Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Solution:
Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).

Question 5.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4,0,0) is equal to 10.
Solution:
Let P(x, y, z) be any point.

The Scalar Triple Product of Vectors in Three-Dimensional Space

Geometrically, the absolute value of the scalar triple product, $iggr vert vec cdot ( vec imes vec ) iggr vert$ is equal to the volume of the parallelepiped formed from the vectors $vec, vec, vec$ as depicted in the following image.

Furthermore, it is important to note that $vec cdot (vec imes vec) = vec cdot (vec imes vec) = vec cdot (vec imes vec)$ , which can be proven by noting the following determinants result in a two row interchanges which is equivalent to multiplying the determinant $vec cdot (vec imes vec)$ by $(-1)(-1) = 1$ .

We're about to look more into lines, planes, and surfaces in general in $mathbb^3$ . The following definition will be important coming up.

 Definition: Three vectors $vec, vec, vec$ are said to be Coplanar if all three vectors lie on the same plane.

For example, the vectors $(0, 1, 2), (0, 2, 3), (0, -1, 3) in mathbb^3$ are coplanar as they all lie on the $yz$ -plane. The following theorem gives us a nice way of checking if three vectors in $mathbb^3$ are coplanar.

• Proof: $Rightarrow$ Suppose that $vec, vec, vec in mathbb^3$ are coplanar. Then these vectors lie on a plane and thus one of the dimensions of the corresponding parallelepiped formed by these vectors has zero length, and so $vec cdot (vec imes vec) = 0$ .
• $Leftarrow$ Suppose that $vec cdot (vec imes vec) = 0$ . Then the volume spanned by $vec, vec, vec$ is zero which implies that these vectors must line on the some similar plane, and thus they are coplanar. $lacksquare$

The Cross Product of Vectors in Three-Dimensional Space

Recall that for any two vectors $vec, vec in mathbb^n$ that the dot product $vec cdot vec = u_1v_1 + u_2v_2 + . + u_nv_n$ produces a scalar. We will now look at another type of vector product known as the Cross Product. We emphasize that the cross product between two vectors is only defined in $mathbb^3$ .

Using the latter definition of the cross product of two vectors in terms of a 3 x 3 Determinant:

Thus, this "determinant" form of the cross product can be rather useful instead of memorizing the longer formula for the cross product. We will now look at some properties of the cross product and prove a few of them. The rest will be left for the reader to prove.

12.2: Vectors in Three Dimensions - Mathematics

This is a basic introduction to the mathematics of vectors. Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law. Forces, displacements, velocities, accelerations, and momenta are examples of physical quantities that may be represented mathematically by vectors. Physical quantities that do not have direction, such as volume, mass or energy, are represented by ordinary number or scalars.

In this article, scalar variables will be denoted as an italicized variable, such as a. Variables that are vectors will be indicated with a boldface variable, such as a, although it is common to see vectors denoted with small arrows above the variable.

In many problems it will be found desirable to resolve a force into two or three components that are perpendicular to each other. These component vectors are generally oriented on a coordinate system, the most popular of which is the two-dimensional Cartesian plane. The Cartesian plane has a horizontal axis which is labeled x and a vertical axis labeled y. One can use the same principle to specify the position of any point in three-dimensional space by three Cartesian coordinates, in which the axes are x, y, and z.

Vectors in multiple-dimension coordinate systems can be resolved into their component vectors. In the three-dimensional case, this results in an x-component, a y-component and a z-component. Figure 1 to the right is an example of a vector, a, resolved into its components, ax, ay and az. When breaking a vector into its components, the vector is a sum of the components,

Adding scalar quantities ignores all information about the directions however, vectors are manipulated somewhat differently &ndash the direction must always be considered. Adding two or more vectors is synonymous to placing the vectors end to end, and creating a new vector running from the starting point to the end point, as demonstrated in Figure 2. If the vectors have the same direction, then this simply means adding the magnitudes, but if they have different directions, the addition is more complex. Vectors are added by resolving them into their components and then adding the components, as shown below,

The subtraction of a vector is defined as the addition of the corresponding negative vector, that is, by a &ndash b we mean a + (&ndashb).

Scalar Multiplication

The operation of multiplying a vector by a scalar is called scalar multiplication. If we define the product na of a scalar n and a vector a as a vector having the same direction as a (if n positive), or a direction opposite to that of a (if n negative), and a magnitude equal to the product of a and of the absolute value of n. The resulting vector is:

n a = n ax + n ay + n az

Scalar multiplication is distributive over vector addition in the following sense:

n (a + b) = n a + n b

We shall, at this point, introduce three vectors of magnitude 1, directed respectively along the positive x-, y- and z-axes. These vectors are called unit vectors and are denoted by i, j and k, respectively. Recalling the definition of the product of a scalar and a vector, we note that the rectangular components ax, ay and az of a vector a may be obtained by multiplying respectively the unit vectors i, j and k by appropriate scalars. We write

a = ax i + ay j + az k

While the scalars ax, ay and az may be positive or negative, depending on the sense of ax, ay and az, their absolute values are respectively equal to the magnitudes of the component vectors ax, ay and az. The scalars ax, ay and az are called the scalar components of the vector a, while the actual components ax, ay and az should be referred to as the vector components of a. However, when there exists no possibility of confusion, the vector as well as the scalar components of a may be referred to simply as the components of a. We note that the scalar component of ax is positive when the vector component ax has the same sense as the unit vector i (i.e. the same sense as the positive x-axis) and negative when ax has the opposite sense. A similar conclusion may be drawn regarding the sign of the scalar components ay and az.

The magnitude or length of the vector a, represented a = |a|, may be obtained by applying the Pythagorean theorem,

a = ( ax 2 + ay 2 + az 2 ) 1/2

The dot product of two vectors a and b is defined as the product of the magnitudes of a and b and the cosine of the angle formed by a and b. The dot product of a and b is denoted by a &bull b and is defined as:

a &bull b = a b cos

Note that the expression just defined is not a vector, but a scalar, which explains why the dot product is also referred to as the scalar product of a and b.

The dot product can also be defined as the sum of the products of the components of each vector,

Combining the above equations, we see that

• The line of action of a × b is perpendicular to the plane containing a and b (see Figure 4).
• The magnitude of a × b is the product of the magnitudes of a and b and the sine of the angle formed by a and b.
• The sense of a × b is such that a person located at the tip of a × b will observe as counterclockwise the rotation through that brings a in line with b. The three vectors a, b and a × b &ndash taken in that order &ndash are said to form a right-handed triad.

The cross product differs from the dot product primarily in that the result of the cross product of two vectors is a vector, thus it is also referred to as the vector product. The cross product is defined as

a × b = a b sin() n

where n is a unit vector perpendicular to both a and b.

The cross product can also be written as

Line of Intersection between two Planes

If we are given two vectors, n1 and n2, that are normal to Plane 1 and Plane 2, then by simple geometrical reasoning, the line of intersection of the two planes is perpendicular to both normals (see Figure 5).

To find the position vector, r, of any point on the line of intersection, we (1) find a vector, v, to which the line is parallel, and (2) find the position vector, a, of a specific point on the line.

As we found above, the cross product of two vectors gives a third vector that is perpendicular to both vectors. We know that v is perpendicular to both n1 and n2 therefore, v is parallel to n1 × n2.

To find a, we find any point on the line of intersection that satisfies the equations of both planes. If we let the point have an x-coordinate of zero, then we have two equations in two unknowns, y and z. Solving for y and z and setting x = 0, we have a point on the line of intersection who's position vector is a.

Substituting v and a into the following yields an equation for the line of intersection:

r = a + t v

Transformation of Coordinates

In astrodynamics it is common to give position vectors in polar coordinates of longitude, l , latitude, b , and radial distance, r. These coordinates are converted to XYZ rectangular coordinates by the following formulae:

x = r cos b cos l
y = r cos b sin l
z = r sin b

Given XYZ coordinates, longitude and latitude are derived as follows:

l = arctan[ y / x ]
b = arctan[ z / (x 2 + y 2 ) 1/2 ] = asin[ z / r ]

where r is simply the vector magnitude. It is important that special care be taken to place l in the correct quadrant.

Three Dimensional Geometry Class 12 Maths Important Questions

CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more.

Three Dimensional Geometry Class 12 Maths Important Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 11 Dimensional Geometry Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Resultant Vector In Three Dimensions

A vector in three-dimensional space. A set of three mutually orthogonal unit vectors Right handed system.

Matching Activity Can You Match The Different Ways Of Describing Movements Across The Grid Math Vector Matching Activity Math Exercises

Rectangular component of a Vector.

Resultant vector in three dimensions. Base vectors for a rectangular coordinate system. So a three-dimensional vector field is given by a function a certain multi-variable function that has a three-dimensional input given with coordinates x y and z and then a three-dimensional vector output that has expressions that are somehow dependent on x y and z Ill just put dots in here for now but well fill this in with an example in just a moment. Scalar multiplication of vectors satisfies the distributive property and the zero vector acts as an additive identity.

To find resultant of forces in 3D We can use R Fx2Fy2Fz2 R 450022250211002 R 2025000050625001210000 R 26522500 R 5150 N. The notation is a natural extension of the two-dimensional case representing a vector with the initial point at the origin and terminal point The zero vector is So for example the three dimensional vector is represented by a directed line segment from point to point Figure. We can draw the vector OP as follows.

Just like two-dimensional vectors three-dimensional vectors are quantities with both magnitude and direction and they are represented by directed line segments arrows. For a vector in three dimensions r xr yr z the magnitude is When multiplying numbers there are three different ways to show that multiplication should be performed. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

Three-dimensional vectors can also be represented in component form. If displacement vectors A B and C are added together the result will be vector R. A is the angle between u and the x-axis.

Suitable for high school physics. Three-dimensional vectors can also be represented in component form. It is the result of adding two or more vectors together.

The 3D vectors are using the x-y-z axes. Now since Resultant is a 3D vector you will not get one angle theta as in case of 2D vector. G is the angle between u and the z-axis.

X or no symbol. With a three-dimensional vector we use a three-dimensional arrow. Vectors can also be multiplied but there are different kinds.

OP sqrt22 32 52 616 units. For example a multiplied by b can be written a x b a. Calculate the angle of three dimensional vectors 3D Vectors with entered vector coordinates.

As shown in the diagram vector R can be determined by the use of an accurately drawn scaled vector addition diagram. The vector vca is drawn as a green arrow with tail fixed at the origin. I review how to find the resultant graphically and then show how to do it algebraically.

Students will be able to solve 3-D particle equilibrium problems by a Drawing a 3-D free body diagram and b Applying the three scalar equations based on one vector equation of equilibrium. A coordinate system represented by base vectors which follow the right-hand rule. The resultant is the vector sum of two or more vectors.

THREE-DIMENSIONAL FORCE SYSTEMS Todays Objectives. The projections of vector A along the x y and z directions are A x A y and A z respectively. V is the angle between u and the y-axis.

Vectors in Two and Three Dimensions Algebra and Trigonometry - James Stewart Lothar Redlin Saleem Watson All the textbook answers and step-by-step explana. These properties of vector operations are valid for three-dimensional vectors as well. Magnitude of a 3-Dimensional Vector.

A vector of unit length. With a three-dimensional vector we use a three-dimensional arrow. You can drag the head of the green arrow with your mouse to change the vector.

A representation of a vector vcaa_1a_2a_3 in the three-dimensional Cartesian coordinate system. We saw earlier that the distance between 2 points in 3-dimensional space is distance AB sqrt x_2-x_12 y_2-y_12 z_2-z_12 For the vector OP above the magnitude of the vector is given by.

The Name That Vector Interactive Is A Skill Building Tool That Presents Users With 12 Vector Addition Chall Precalculus Precalculus Activities Physics Concepts

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12.2: Vectors in Three Dimensions - Mathematics

Values used in physics are either scalar or vector.

Scalar values indicate positive magnitudes such as mass, speed, work, charge, or energy, and have no directional component.

Many values do, however, have a directional component. These includes position, velocity, acceleration, and force, which all require a vector to be described properly. See interactive diagram above.

For a three-dimensional vector, we can use three numbers to describe the end-point of a directed line segment that starts at the origin, with each number indicating the displacement along each of the three axis, the x axis, y axis, and z axis.

We can determine the length of such a vector with the equation:

length = sqrt(Rx2 + Ry2 + Rz2)

We can describe the direction of the vector in terms of two angles, where:

We can think of the vector R in the diagram as the sum of the unit vectors, each multiplied by the magnitude of R in each respective dimension (see green dots), or


text:
Short-term wholesale funding, which acted as an accelerant during the financial crisis, contributes the most in the Federal Reserve’s setting of banks’ systemic risk scores.

Almost a third of the average aggregated scores for the eight current US global systemically important banks (G-Sibs) over the last six quarters were attributable to the short-term wholesale funding indicator, Risk Quantum analysis shows.

The indicator is one of five used by the Fed to gauge banks’ systemic riskiness under

#### 7 days in 60 seconds

###### CCPs, handling default, and cyber risk

The week on Risk.net, September 15–21, 2018

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text:
# Measuring An Angle

Today we will focus on a topic from elementary geometry namely the concept of measurement of angles. The idea of an angle is a simple one in the sense that it is made by two rays emanating from the same point. But the measurement of angles is not that simple as it appears. Many theorems in elementary geometry deal with ideas which involve the concept of measurement of angles but they assume the understanding of this measurement in an implicit fashion.

### Measuring Angles

For example we have the properties of a transversal cutting two parallel lines and the corresponding as well as alternate angles being equal. What do we mean when we say that the two corresponding angles (in the last statement) are equal? Well luckily the concept of equality (or congruence) of geometrical shapes is much easier to define and comprehend. Two shapes are equal if they can be exactly superimposed on one another. Thus for two angles to be equal we need to note that when we superimpose one angle on the other the vertices as well as the rays of both the angles must coincide exactly.

In the same manner by superimposing angles such that the vertices and at least one of the rays of both angles coincide we can compare two angles and determine which of the angles is greater or smaller. This is similar to comparing two line segments by the use of a divider. However it turns out that practical applications of geometry not only require us to compare line segments but rather also measure them in a precise manner. Similar is the case with angles. A lot of practical applications of geometry (and more properly trigonometry) require us to measure angles in a precise manner.

Naturally measuring anything requires two fundamental ideas:
• Defining a standard unit for measurement
• An additive law defining how any magnitude of the "entity being measured" can be expressed in terms of the standard unit.
For example in case of line segments we can take any specific line segment $OA$ and define its measure (or length) to be $1$ and then the length of any other line segment $PQ$ must be seen in comparison with the length of standard line segment $OA$ whose length we have defined to be $1$. More technically if line segment $PQ$ has a point $R$ in between $P$ and $Q$ such that the line segment $OA$ can be superimposed exactly with both the line segments $PR$ and $RQ$ then we say that the line segment $PQ$ has length $2$. The extension to rational lengths is made possible by a very standard construction of dividing the unit line segment $OA$ into a finite number of line segments of equal length. Finally the extension to irrational lengths is an assumption which links the branches of elementary geometry and real analysis.

The point of discussion in previous paragraph is to emphasize the fact that the concept of length of a line segment is a very fundamental one in terms of measuring magnitudes of geometrical shapes. Once we have the grasp of length it is possible to define the concept of areas of geometrical shapes (at least those shapes which are made up of line segments only). Extension to lengths and areas of shapes consisting of general curves is made possible via real analysis (via the concept of a definite integral). For the present discussion we will assume this extension of lengths and areas for any geometrical shape and proceed to measure angles.

#### Measuring Angles: Step 1

Intuitively the way we visualize an angle and think about comparison of angles, it is almost obvious that a greater angle somehow signifies that its rays are farther apart from one another compared to those of a smaller angle. Thus the measurement of an angle must be related somehow to the spacing between its rays. But again it is obvious that as we move away from the vertex of an angle the rays look diverging apart from each other more and more. Hence to concretely measure the spacing between the rays it is important to fix a distance from the vertex at which we want to measure the spacing between two rays.

So let $POQ$ be an angle which we wish to measure and let $A, B$ be points on rays $OP, OQ$ respectively such that the length of line segments $OA$ and $OB$ is $1$. An intuitive measure of $\angle POQ$ must be linked with a measure of separation of the points $A$ and $B$. What could be simpler than just measuring the length of line segment $AB$? So let us define the measure of $\angle POQ$ to be the length of line segment $AB$ where $OA = OB = 1$.
 Measuring an Angle: Step 1
An equally simple measure of $\angle POQ$ could be the area of $\Delta AOB$. In fact using area would really give a measure of space between the rays $OP$ and $OQ$ and the line segment $AB$.

However if we observe carefully both the definitions (either using length of $AB$ or area of $\Delta AOB$) suffer from two defects:
• The comparison of angles fails miserably, and
• The additive property also fails in a very obvious manner
The failure of comparison of angles (using the area of $\Delta AOB$ as a measure of $\angle POQ$) is shown via the following figure:
 Failure of Comparison of Angles
In the above figure we have $OA = OB = OC = 1$ and it is obvious that the $\angle ROQ$ is significantly larger than the $\angle POQ$, but the area of $\Delta COB$ is much less than the area of $\Delta AOB$. However note that had we used the length $AB$ as measure of $\angle POQ$ then the comparison would have remained valid in the above figure because length of $BC$ is obviously greater than that of $AB$. In order to see the failure of length $AB$ as a measure of $\angle POQ$ we need to understand that associated with a configuration of two rays of $OP$ and $OQ$ emanating from same point $O$ there are two angles possible: one the interior angle (smaller one) between $OP$ and $OQ$ and another the exterior angle which is the portion of plane outside the rays $OP$ and $OQ$ and which is significantly larger than the interior angle. When we consider this view it is obvious that both these angles must be measured by same length $AB$. And it might turn out that a significantly larger angle has a comparatively smaller measure.

The additive property of measurement of angles clearly does not hold with either of these definition based on length (of $AB$) and area (of $\Delta AOB$) as evident from the figure below:
The length $BC$ is clearly not equal to the sum of lengths of $AB$ and $AC$, rather it is slightly less than this sum. However from the same figure above it is obvious that the measure of $\angle ROQ$ must be equal to the sum of measures of $\angle ROP$ and $\angle POQ$. The same problem is seen with the areas of the triangles also as the area of $\Delta COB$ is slightly less than the sum of areas of $\Delta COA$ and $\Delta AOB$.

#### Measuring Angles: Step 2

What went wrong with the previous definition of measure of angles? What else we need to associate with the gap between two rays emanating from the same point? Well there is no other suitable choice than the previous two approaches based on length of a line segment and area of a triangle. But they suffer from obvious defects as seen in previous paragraphs. Is it possible to fix these defects? Let us give it a try and focus on the defects more closely and perhaps we might be able to fix them.

Let's take the issue of additivity. We find that the additivity fails by a very small margin. Consider the following figure:
If we compare this figure with the previous figure then we note that the additivity fails here by an even smaller margin. To be explicit the difference between the length $BD$ and the sum of lengths of $AB, AC$ and $CD$ is very small. Again the difference between area of $\Delta DOB$ and sum of areas of $\Delta DOC, \Delta COA$ and $\Delta AOB$ is very small. What is so different about this figure compared to the last figure? Here one angle has been expressed as the sum of three smaller angles whereas in previous case an angle was expressed as sum of two smaller angles. Intuitively it is obvious that if we express an angle as a sum of 4 smaller angles then the margin of failure of additivity in the measurement of angles will be even less. Thus to fix the issue of additivity we need to express an angle as a sum of as many smaller angles as possible.

Leaving aside the practical problem of drawing an angle as a sum of say $100$ smaller angles, we can see that this is the way forward. Thus from a theoretical point of view the best option would be to let the number of smaller angles tend to infinity. When the number of smaller angles tends to infinity then we see that the polygonal line $BACD$ turns into an arc of a circle of radius $1$ and centre $O$. And the region $OBACD$ becomes the corresponding sector associated with the arc. The situation is illustrated in the figure below:
 Measuring an Angle as Sum of Many Many Angles
Starting from an intuitive approach we have finally reached the modern definition of the measure of an angle. If $POQ$ is an angle then its measure is defined to be the length of arc $AB$ of a circle centered at $O$ and radius $1$ such that $A$ lies on ray $OP$ and $B$ lies on ray $OQ$. The measure of $\angle POQ$ can also be defined as twice the area of sector $AOB$. The unit of measurement of angles as defined above is called a radian and there is no specific symbol used for this unit of measurement.
 Measure of an Angle
The factor of $2$ comes because of a specific relation between the circumference of a unit circle ($2\pi$) and its area ($\pi$) which is based on a fact of integral calculus namely $$\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}} = 2\cdot\int_{0}^{1}\sqrt{1 - x^{2}}\,dx = \frac{\pi}{2}$$ where the integral on the left represents the length of a quadrant of a unit circle and the integral on the right represents the area of corresponding sector. We thus see that measuring an angle is intimately connected to measuring stuff related to a circle. No wonder we used semi-circular protractors in school to measure angles in degrees!

#### 8 comments :: Measuring An Angle

1. Hello i am godonichia from stackexchange .I have seen your blog and was fascinated by your interest in maths exp Ramanujan's area . I have too a passion for it .Can you guide me regarding books which are preps to entering into number theory .I have studied basic single,multivariable calculus,linear algebra,ODE,.Hoping to get reply soon .Thanks

2. @Godonicha Augustus,
I have given some references at http://paramanands.blogspot.com/p/bibliography.html These books have helped me in writing a lot of these blog posts. You should read these. They are good for self study.

Regards,
Paramanand

3. @Paramanand Singh Can you suggest a order for doing these books in order to understand Ramanujan Notebooks .Assume usual high school mats knowledge .Thanks

4. @Ritesh Arora,
For every reader of my blog I would first suggest to read "A Course of Pure Mathematics" completely. This books forms the basis of all that is necessary to understand rigorous calculus/real analysis. Then you can go for "An Elementary Treatise on Elementary Functions" and finally to Ramanujan Notebooks. You can also starts with the blogs posts "Pi and the AGM" (full series - this was the first topic on my blog) followed by "Magic of Theta Functions", Elliptic Functions (full series), Elementary Approach to Modular Equations by both Jacobi and Ramanujan. This will set you up with Ramanujan stuff. Then you can read further posts on Modular Equation and Rogers-Ramanujan.

5. -my comment is from France
it is not on this particular subject "Measuring An Angle"
but for to indicate to Paramanand Singh
that on his 107 (at this date) pdf files of these Paramanand's Math Notes
--but one (only) is not readable, it is:
2011-05-17 Irrationality of p(PI) Lambert's Proo
(...litte problem...)
-about his "Paramanand's Math Notes": fabulous
-thanks

Anonymous

6. @Anonymous (from France)
The issue with the PDF file related to "Irrationality of $\pi$(PI): Lambert's Proof" has been fixed now. The link for the PDF file had a small typo and hence it was not showing the right PDF.

Regards,
Paramanand


text:
# Solving Recurrences via Characteristic Polynomial with Imaginary Roots

In algorithm analysis you often have to solve recurrences. In addition to Master Theorem, substitution and iteration methods, there is one using characteristic polynomials.

Say I have concluded that a characteristic polynomial $x^2 - 2x + 2$ has imaginary roots, namely $x_1 = 1+i$ and $x_2 =1-i$. Then I cannot use

$\qquad c_1\cdot x_1^n + c_2\cdot x_2^n$

to obtain the solution, right? How should I proceed in this case?

• Welcome! Note that you can use LaTeX by $...$. – Raphael Mar 31 '12 at 10:38
• I am confused. I am sure you mean the method using characteristc polynomials, not equations. What is $j$? The solutions of the equation you give are not imaginary, but merely irrational. What do you mean by "apply the [polynomial]"? – Raphael Mar 31 '12 at 10:40
• He's adopted the physicist's habit of misspelling $i$. – JeffE Mar 31 '12 at 11:41
• Of course, you can actually. First, the solution satisfies the reoccurrence. Second, the solution space is of dimension 2. – Strin Apr 3 '12 at 5:23

Yes, the solution is in fact $T(n) = \alpha(1+i)^n + \beta(1-i)^n$ for some constants $\alpha$ and $\beta$ determined by the base cases. If the bases cases are real, then (by induction) all the complex terms in $T(n)$ will cancel, for all integer $n$.
For example, consider the recurrence $T(n) = 2T(n-1) - 2T(n-2)$, with base cases $T(0)=0$ and $T(1)=2$. The characteristic polynomial of this recurrence is $x^2-2x+2$, so the solution is $T(n) = \alpha(1+i)^n + \beta(1-i)^n$ for some constants $\alpha$ and $\beta$. Plugging in base cases gives us $$T(0) = \alpha(1+i)^0 + \beta(1-i)^0 = \alpha+\beta = 0\\ T(1) = \alpha(1+i)^1 + \beta(1-i)^1 = (\alpha+\beta) + (\alpha-\beta)i = 2$$ which implies $$\alpha + \beta = 0 \\ \alpha - \beta = -2i$$ which implies $\alpha = -i$ and $\beta = i$. So the solution is $$T(n) = i\cdot ((1-i)^n - (1+i)^n).$$
This function oscillates between $\sqrt{2}^n$ and $-\sqrt{2}^n$ with a "period" of 4. In particular, we have $T(4n) = 0$ for all $n$, because $(1-i)^4 = (1+i)^4 = -4$ (and because I chose the base case $T(0)$ carefully).
• Not necessarily. If the roots of the characteristic function are $2$, $1+i$, and $1-i$, for example, the function will oscillate around $\alpha 2^n$ for some $\alpha$, but (with appropriate base cases) it will always be positive. – JeffE Mar 31 '12 at 13:24


text:
# Lagrangian for fields AND particles?

1. Jan 7, 2010

### pellman

In general what does a Lagrangian for a system consisting of interacting fields and particles look like?

It can't be, for example,

$$L=\sum{\frac{1}{2}mv_j^2+A(x_j)\inner v_j}$$

That would be for a system of particles in a fixed, i.e. "background", field. I'm interested in how we can mix particles and fields in Lagrangian mechanics. I know how to write down, as above, the Lagrangian for particles influenced by a field. And I know how to write down a Lagrangian (density) for a field with fixed (continuum) sources. But what does a Lagrangian (density?) that governs both fields and discrete sources look like?

No need to lay out the most general case. Just a simple example will suffice.

2. Jan 9, 2010

### pellman

Ok. No replies. I can take this now to the next step myself. Then maybe someone else can help from there.

Supposedly, the full action for both EM fields and (dynamic) sources is

$$-m \int d\tau \sqrt{- g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} + q \int dx'^4 \int d\tau ~ \delta^4(x'-x(\tau)) \frac{dx^\mu (\tau)}{d\tau} A_\mu - \frac{1}{4} \int d^4 x F^{\alpha \beta} F_{\alpha \beta}$$


text:
# Graph y=3*5^x

y=3⋅5x
Exponential functions have a horizontal asymptote. The equation of the horizontal asymptote is y=0.
Horizontal Asymptote: y=0
Graph y=3*5^x


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# Principles of Math MPM2D

This course enables students to broaden their understanding of relationships and extend their problem-solving and algebraic skills through investigation, the effective use of technology, and abstract reasoning. Students will explore quadratic relations and their applications; solve and apply linear systems; verify properties of geometric figures using analytic geometry; and investigate the trigonometry of right and acute triangles. Students will reason mathematically and communicate their thinking as they solve multi-step problems. Prerequisite: Grade 9 Principles of Math MPM1D

# Analytic Geometry

## Given the equation for a circle with a center at (h, k)...

\begin{align} (x - h)^2 + (y - k)^2 = r^2 \\ \\ \end{align}

## A volcano in Iceland erupts with center C(10, 14), the ash cloud formed has radius 3 km and can be modeled with the circle equation below. A weather drone plane flies on route y = 0.8x + 9.

\begin{align} (x - 10)^2 + (y - 14)^2 = 3^2 \\ \\ \end{align}

# Geometric Properties

## Given the following points, determine the circumcenter of the triangle. (Hint: a circumcenter is where all the right bisectors intersect.)

L(2, 2)     M(1, 1)     N(0, -3)

A (5, 9)
B (-10, -9)
C (15, -10)

xa × xb = xa + b

xa ÷ xb = xa - b

(xa)b = xa × b

## Simplify fully to an integer or fraction using positive exponents, showing your work without using a calculator.

$a^{-n} = \left( \cfrac{1}{a^{\ +n}} \right) \quad\quad or \quad\quad \left( \cfrac{a}{b} \right)^{\!\!-n} = \left( \cfrac{b}{a} \right)^{\!\!+n}$

## Solve the following exponential equations, similar to the steps shown in the example below. Simplify fully.

\begin{align} 3^{(3x + 2)} & = 243 \\ 3^{(3x + 2)} & = 3^{(5)} \\ 3x + 2 & = 5 \\ 3x & = 5 - 2 \\ 3x & = 3 \\ x & = 1 \\ \\ \end{align}

## The table below shows the flight path of a ball, with distance and height in meters.

 Distance (x) Height (y) 0 1 1 8 2 13 3 16 4 17 5 16 6 13 7 8 8 1

## A basketball shot is taken from a horizontal distance of 5 m from the hoop. The height of the ball can be modeled by the equation below. Where h is height in meters, and t is time in seconds since the ball was released.

$h(t) = -7.3t^2 + 8.25t + 2.1$

## Given the general, vertex form of a quadratic.

$y = ±\,a(x - h)^2 + k$

## Given the general form of a quadratic, determine the equation of a quadratic function with the transformations listed below....

$y = a(x - h)^2 + k$

## Based on the value 'n' determine some of the horizontal transformations below.

$y = a\left[ n\left(x - h\right)\right]^2 + k$

## A quadratic function passes through the points (-1, 1), (0, -3), and (5, 1).

### Determine the equation in vertex form. Solution Hint Clear Info $\dfrac{\quad\quad}{\quad\quad} \Bigg(x \quad\quad\quad \Bigg)^2 - \dfrac{\quad\quad}{\quad\quad}$ Incorrect Attempts: CHECK Hint Unavailable First determine the Axis of Symmetry (AOS) of the quadratic. The AOS goes through the vertex and is the midpoint of two points on the same horizontal line: (-1, 1) and (5, 1). \begin{align} \text{AOS} & = \text{Midpoint of x} \\ \\ x & = \dfrac{x_1 + x_2}{2} \\ \\ x & = \dfrac{-1 + 5}{2} \\ \\ x & = 2 \\ \\ \end{align} Substitute AOS and the easiest point (0, -3) into the vertex form... This AOS represents 'h' in the equation... \begin{align} y & = a(x - h)^2 + k \\ \\ -3 & = a(0 - 2)^2 + k \\ \\ -3 & = 4a + k \\ \\ \end{align} Substitute AOS and any other point, (-1, 1)... \begin{align} 1 & = a(-1 - 2)^2 + k \\ \\ 1 & = 9a + k \\ \\ \end{align} Solve with elimination... \begin{align} & -3 = 4a + k \\ (-) \ & \underline {1 = 9a + k} \\ \\ & -4 = -5a \\ \\ & a = \dfrac{4}{5} \\ \\ \end{align} Now find the last value, 'k by subbing in the last point (5, 1)... \begin{align} y & = a(x - h)^2 + k \\ \\ y & = \dfrac{4}{5}(x - 2)^2 + k \\ \\ 1 & = \dfrac{4}{5}(5 - 2)^2 + k \\ \\ 1 & = \dfrac{36}{5} + k \\ \\ k & = -\dfrac{31}{5} \\ \\ \end{align} Here is your equation in vertex form, \begin{align} y & = \dfrac{4}{5}(x - 2)^2 - \dfrac{31}{5} \\ \\ \end{align}

Specific Topic General Topic School Date
The Correct Way to Expand a Squared Binomial Expanding Branksome Sep 2013
Factoring a Perfect Square Trinomial Quadratics Branksome Sep 2013
Factoring Unordered Trinomials Example 1 Quadratics Branksome Sep 2013
Factoring Unordered Trinomials Example 2 Quadratics Branksome Sep 2013

## Now that you understand more about factoring it's time to take the training wheels off - no more decomposition (aka splitting and grouping). Factor straight into the parenthesis like the example shown below.

$6x^2 + 11x + 3$ What multiplies to +18 and adds to +11? = +9 and +2 \begin{align} & = (2x \quad\quad )(3x \quad\quad ) \\ \\ & = ( \ \underbrace{2x + \underbrace{3 \ )( \ 3x}_{+9} + 1 }_{+2} \ ) \\ \\ \end{align}

## Factor the following perfect squaresfully, using any method (direct is encouraged). The general formulas for perfect squares are given below.

\begin{align} & a^2 + 2ab + b^2 \\ \\ & = (a)^2 + 2ab + (b)^2 \\ \\ & = (a + b)^2 \\ \\ \end{align} \begin{align} & a^2 - 2ab + b^2 \\ \\ & = (a)^2 - 2ab + (b)^2 \\ \\ & = (a - b)^2 \\ \\ \end{align}

## Given the general formula for a perfect square quadratic...

E.g. 1) (a)2x2 + 2(a)(b)x + (b)2
= (ax + b)2
E.g. 2) x2 + 6x + 9
= (a)2x2 + 2(a)(b)x + (b)2
= (1)2x2 + 2(1)(3)x + (3)2
= (1x + 3)2

## Factor the difference of squares fully (meaning common factor whenever possible). Show as much work in your notes as you need.

\begin{align} & 1x^2 - 9 \\ \\ & = (1x + 3)(1x - 3) \\ \\ \end{align} \begin{align} & 75a^2 - 27 \\ \\ & = 3(25a^2 - 9) \\ \\ & = 3\left[ (5a)^2 - (3)^2 \right] \\ \\ & = 3(5a + 3)(5a - 3) \\ \\ \end{align}

## A triangular sandbox is surrounded by a rectangular field of grass. The dimensions of the sandbox and field are listed below, in terms of x.

 Base (cm) Height (cm) Length (cm) Width (cm) Triangle 2x x + 4 Rectangle x + 1 4x - 4

### Determine the most simplified expression for the total surface area of a cube with side lengths equal to (2x - 3). Solution ⁰¹²³⁴⁵⁶⁷⁸⁹⁻⁺⁽⁾₀₁₂₃₄₅₆₇₈₉₋₊₍₎ Incorrect Attempts: CHECK Hint Unavailable There are 6 equal faces on a cube.The area of each face is (2x - 3)(2x - 3) = (2x - 3)2 The total surface area = 6(area) = 6(2x - 3)2

Specific Topic General Topic School Date
Solving for X by Completing the Square Factoring Quadratics Branksome Sep 2013

## Given the equation already in factored form.

\begin{align} y = x(2x - 3) \\ \\ \end{align}

## The revenue, expense, and profit functions are given below, where x is the number of items sold.

\begin{align} & Revenue = 5x^2 + 200 \\ \\ & Expense = 35x^2 - 130x + 100 \\ \\ & Profit = Revenue \ - \ Expense \end{align}

## The cost of operating a machine, in dollars, is given by the formula below where t is time, in hours, that the machine operates.

C(t) = 2t2 - 16t + 682

Base = x
Height = x + 7
Hypotenuse = 17

# Trigonometry with Acute and Obtuse

## Given the complex shape, determine the missing quantities stated below.

### Explain how the Cosine law is based on the Pythagorean theorem. [3] Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Pythagorean Theorem $a^2 + b^2 = c^2$ Cosine law when angle (A˚) = 90˚ \begin{align} a^2 & = b^2 + c^2 - 2bc \cos A˚ \\ \\ a^2 & = b^2 + c^2 - 2bc \cos (90˚) \\ \\ a^2 & = b^2 + c^2 - 2bc \cancelto{0}{\cos (90˚)} \\ \\ a^2 & = b^2 + c^2 - \cancelto{0}{2bc (0)} \\ \\ a^2 & = b^2 + c^2 \\ \\ \end{align} The Cosine law is the Pythagorean theorem when the angle is 90˚...

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text:
# Time period of two blocks

1 vote
44 views

For the system shown in the figure initially spnngs are in natural length (natural length
of both are same). Two blocks ‘A’ and 'B’ of equal mass 'M' are released from rest
simultaneously The contact between the string and pulley is frictionless and pulley is
fixed. The masses of string and springs are negligible. How the time period of oscillation of both blocks are same.

In this force in both springs would be always same .
That is kx$_1$ =2kx$_2$ .

But now how can we comment on their time period .com

Have you written and solved the equations of motion?

Do you think periods will be the same or different for the 2 masses? Why?
http://i.imgur.com/lFXdEe2.gif
Don't know how to write equation of motion in spring.

1 vote

The question states that the pulley is frictionless and fixed but not that it is massless. I think the latter (fixed) must be a mistake : the pulley must be free to rotate, in order to get the result that the blocks have the same period. We must also assume that the pulley is massless. Then tension in the string and springs is always the same on both sides of the pulley, as is the case when there is string but no springs.

The masses are equal and the tensions acting on them are equal, so the acceleration must also be equal : $m\ddot x=mg-T$ (taking down as +ve for both masses). Both masses have the same equation of motion, so the oscillation frequency and period is the same.

It is tempting to write $T=kx_1=2kx_2$ where $x_1, x_2$ are the position co-ordinates of the masses. Then we seem to have different equations of motion and different frequencies :
$\ddot x_1+\frac{k}{m}x_1=g$
$\ddot x_2+\frac{2k}{m}x_2=g$.
However, this is incorrect. It assumes that the upper ends of the springs are fixed in position. They are not. The pulley turns freely. Then some of the larger extension of spring 1 is transferred smoothly and instantaneously to the other side, compensating for the smaller extension of spring 2. Therefore the extension of spring 1 is not equal to the displacement of mass 1, and likewise for spring 2/mass 2.

The masses are released at the same time, and the above mechanism keeps the lower ends of the springs always level, where the masses are attached. So the masses oscillate in phase.

It is a little more difficult to anticipate what happens if the masses are not released at the same time...

answered Feb 13, 2017 by (28,746 points)
selected Mar 8, 2017 by koolman
Can we say bot the springs have same acceleration at any instant ,so the time period is also same .
The accelerations of the *blocks* are the same, therefore their periods are the same. Different parts of the springs accelerate at different rates, because they expand.


text:
# nth order differentiation challenge

#### MarkFL

Staff member
Let:

$$\displaystyle f(x)=x\sin(x)$$

Derive a formula for:

$$\displaystyle f^{(n)}(x)$$

Using this, infer a formula for:

$$\displaystyle \frac{d^n}{dx^n}\left(x\cos(x) \right)$$

edit: I wanted to make sure it is clearly understood that:

$$\displaystyle f^{(n)}(x)\equiv\frac{d^n}{dx^n}\left(f(x) \right)$$

#### Ackbach

##### Indicium Physicus
Staff member
Using the Leibnitz rule for the $n$-th derivative of a product, which in general is
$$\frac{d^{n}}{dx^{n}}[f(x)\,g(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}f(x)}{dx^{j}} \frac{d^{n-j}g(x)}{dx^{n-j}}\right],$$
we have that
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}x}{dx^{j}} \frac{d^{n-j} \sin(x)}{dx^{n-j}} \right].$$
Now, there are only two non-zero terms: $j=0$ and $j=1$. Hence, the sum collapses down to those two terms:
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= {n \choose 0} x \frac{d^{n} \sin(x)}{dx^{n}}+{n \choose 1} \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
Note that
$${n \choose 0}=1,\quad \text{and} \quad {n \choose 1}=n,$$
so we further simplify as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \frac{d^{n} \sin(x)}{dx^{n}}+n \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
To finish, we can use the fact that
$$\frac{d^{n}}{dx^{n}} \sin(x)= \sin \left( \frac{n \pi}{2}+x \right),$$
to write the derivative as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \sin \left( \frac{n \pi}{2}+x \right)+n \sin \left( \frac{(n-1) \pi}{2}+x \right).$$

The same formula happens to work for $\cos$, so simply replace $\sin$ with $\cos$ to get the equivalent formula for $(d^{n}/dx^{n})(x \cos(x))$.

#### MarkFL

Staff member
Using the Leibnitz rule for the $n$-th derivative of a product, which in general is
$$\frac{d^{n}}{dx^{n}}[f(x)\,g(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}f(x)}{dx^{j}} \frac{d^{n-j}g(x)}{dx^{n-j}}\right],$$
we have that
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}x}{dx^{j}} \frac{d^{n-j} \sin(x)}{dx^{n-j}} \right].$$
Now, there are only two non-zero terms: $j=0$ and $j=1$. Hence, the sum collapses down to those two terms:
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= {n \choose 0} x \frac{d^{n} \sin(x)}{dx^{n}}+{n \choose 1} \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
Note that
$${n \choose 0}=1,\quad \text{and} \quad {n \choose 1}=n,$$
so we further simplify as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \frac{d^{n} \sin(x)}{dx^{n}}+n \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
To finish, we can use the fact that
$$\frac{d^{n}}{dx^{n}} \sin(x)= \sin \left( \frac{n \pi}{2}+x \right),$$
to write the derivative as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \sin \left( \frac{n \pi}{2}+x \right)+n \sin \left( \frac{(n-1) \pi}{2}+x \right).$$

The same formula happens to work for $\cos$, so simply replace $\sin$ with $\cos$ to get the equivalent formula for $(d^{n}/dx^{n})(x \cos(x))$.

I recall with fondness "discovering" and proving by induction the formula you used up front when fiddling with reduction of order techniques in my ODE course.

I will give others time before posting my own solution.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Another way:

Consider $h:\mathbf{R}\to \mathbf{C},\;\; h(x)=x\cos x+ix\sin x=xe^{ix}$. Then, $$\frac{d^n}{dx^n}(x\cos x)=\text{Re }\left(h^{(n)}(x)\right)\;,\quad \frac{d^n}{dx^n}(x\sin x)=\text{Im }\left(h^{(n)}(x)\right)$$

#### Ackbach

##### Indicium Physicus
Staff member

I recall with fondness "discovering" and proving by induction the formula you used up front when fiddling with reduction of order techniques in my ODE course.

I will give others time before posting my own solution.
And I remember taking a graduate-level physics course where I was the only mathematical physics guy there, and being asked to put a proof of this formula on the board. The prof asked me to include every step, including the formula for the entry in Pascal's Triangle as the sum of the two numbers directly above it. (The prof seemed to think that the rest of the students couldn't get this on their own - I suppose maybe because he had seen their homework!) I remember thinking to myself that this really wasn't that big a deal. The other students should have been able to do it. Indeed, their physics abilities were far beyond my own. But in the words of the prof, "They don't know how to differentiate!"

My consolation is that all theoretical physicists these days are mathematicians - they have to be, because the math people have gotten so far behind the physics people, that whenever the physicists look around for a needed tool, it's no longer there like it was, say, even 70 years ago.

#### anemone

##### MHB POTW Director
Staff member
My approach:

We're given $$\displaystyle f(x)=x\sin(x)$$

I first find the first few nth derivative of the function of $$\displaystyle f(x)=x\sin(x)$$ to see if there is any pattern to be observed...

$$\displaystyle f^{(1)}(x)=x\cos x +\sin x$$
$$\displaystyle f^{(2)}(x)=-x\sin x +2\cos x$$
$$\displaystyle f^{(3)}(x)=-x\cos x -3\sin x$$
$$\displaystyle f^{(4)}(x)=x\sin x -4\cos x$$
$$\displaystyle f^{(5)}(x)=x\cos x +5\sin x$$
$$\displaystyle f^{(6)}(x)=-x\sin x +6\cos x$$...

At this point, it is easy to generate the general formula for $$\displaystyle f^{(n)}(x)$$ where

$$\displaystyle f^{(n)}(x)=x\sin (\frac {\pi n}{2}+x)-n\cos (\frac {\pi n}{2}+x)$$

and it can easily be shown to be true using induction method.

Let $$\displaystyle P(n)$$ be the statement $$\displaystyle f^{(n)}(x)=x\sin (\frac {\pi n}{2}+x)-n\cos (\frac {\pi n}{2}+x)$$.

$$\displaystyle P(1)$$ asserts that $$\displaystyle f^{(1)}(x)=x\sin (\frac {\pi }{2}+x)-\cos (\frac {\pi }{2}+x)=x\cos x +\sin x$$ which is clearly true.

Next, suppose $$\displaystyle P(n)$$ is true for $$\displaystyle n=k$$. We need to prove $$\displaystyle P(n)$$ is true for $$\displaystyle n=k+1$$.

For $$\displaystyle P(k+1)$$, we have

$$\displaystyle f^{(k+1)}(x)=\frac{d}{dx} \left(x\sin (\frac {\pi k}{2}+x)-k\cos (\frac {\pi k}{2}+x)\right)$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=x\cos (\frac {\pi k}{2}+x)(1)+(1)(\sin (\frac {\pi k}{2}+x)-(k)(-\sin (\frac {\pi k}{2}+x))$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=x\cos (\frac {\pi k}{2}+x)(1)+(1)(\sin (\frac {\pi k}{2}+x)+k\sin (\frac {\pi k}{2}+x)$$

thereby showing that indeed $$\displaystyle P(k+1)$$ holds and this finished our work.

Now, we have $$\displaystyle f(x)=x\cos(x)$$.

Notice that there is a phase shift (delay) of $$\displaystyle \frac{\pi}{2}$$rad from $$\displaystyle y=x\sin x$$ to $$\displaystyle y=x\cos x$$ and also the signs of the two terms are of the opposite have been shifted in the other direction by $$\displaystyle -\frac{\pi}{2}$$ rad.

Therefore, the general formula for $$\displaystyle f^{(n)}(x)$$ if $$\displaystyle f(x)=x\cos x$$ can be obtained from:

$$\displaystyle f^{(n)}(x)=-x\sin (\frac {\pi n}{2}+x-\frac{\pi}{2})+n\cos (\frac {\pi n}{2}+x-\frac{\pi}{2})$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;=x\cos (\frac {\pi n}{2}+x)+n\sin (\frac {\pi n}{2}+x)$$

#### MarkFL

Staff member
Good work, anemone!

One of my approaches is quite similar to that which you gave, the other being that given by Ackbach.

This is how I formulated this approach:

Given:

$$\displaystyle f(x)=x\sin(x)$$

we find:

$$\displaystyle f'(x)=x\cos(x)+\sin(x)$$

$$\displaystyle f''(x)=-x\sin(x)+2\cos(x)$$

$$\displaystyle f'''(x)=-x\cos(x)-3\sin(x)$$

At this point, we may state the hypothesis:

$$\displaystyle f^{(n)}(x)=x\sin\left(x+n\frac{\pi}{2} \right)-n\cos\left(x+n\frac{\pi}{2} \right)$$

Having already demonstrated the base case (and the next few), we may use as our inductive step, differentiation of $P_n$ with respect to $x$:

$$\displaystyle f^{(n+1)}(x)=x\cos\left(x+n\frac{\pi}{2} \right)+(n+1)\sin\left(x+n\frac{\pi}{2} \right)$$

Now, using the identities:

(1) $$\displaystyle \sin\left(\theta-\frac{\pi}{2} \right)=-\cos(\theta)$$

(2) $$\displaystyle \cos\left(\theta-\frac{\pi}{2} \right)=\sin(\theta)$$

we may write:

$$\displaystyle f^{(n+1)}(x)=x\sin\left(x+(n+1)\frac{\pi}{2} \right)-(n+1)\cos\left(x+(n+1)\frac{\pi}{2} \right)$$

We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction.

Now, if we are given:

$$\displaystyle g(x)=x\cos(x)=x\sin\left(x+\frac{\pi}{2} \right)$$

Then from the above, we may infer:

$$\displaystyle g^{(n)}(x)=x\sin\left(x+(n+1)\frac{\pi}{2} \right)-n\cos\left(x+(n+1)\frac{\pi}{2} \right)$$

and using (1) and (2), we find:

$$\displaystyle g^{(n)}(x)=x\cos\left(x+n\frac{\pi}{2} \right)+n\sin\left(x+n\frac{\pi}{2} \right)$$

Another way to infer this second formula is to write:

$$\displaystyle h(x)=x\sin\left(x+\frac{\pi}{4} \right)$$

Thus, we know:

$$\displaystyle h^{(n)}(x)=x\sin\left(x+\frac{\pi}{4}+n\frac{\pi}{2} \right)-n\cos\left(x+\frac{\pi}{4}+n\frac{\pi}{2} \right)$$

Using the identities:

$$\displaystyle \sin\left(\theta+\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}\left(\sin(\theta)+\cos(\theta) \right)$$

$$\displaystyle \cos\left(\theta+\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}\left(\cos(\theta)-\sin(\theta) \right)$$

we may write:

$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(x\left(\sin \left(x+n\frac{\pi}{2} \right)+\cos\left(x+n\frac{\pi}{2} \right) \right)-n \left(\cos\left(x+n\frac{\pi}{2} \right)-\sin \left(x+n\frac{\pi}{2} \right) \right) \right)$$

$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(x\sin \left(x+n\frac{\pi}{2} \right)-n\cos\left(x+n\frac{\pi}{2} \right)+x\cos\left(x+n\frac{\pi}{2} \right)+n\sin \left(x+n\frac{\pi}{2} \right) \right)$$

$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(\frac{d^n}{dx^n}\left(x\sin(x) \right)+x\cos\left(x+n\frac{\pi}{2} \right)+n\sin \left(x+n\frac{\pi}{2} \right) \right)$$

Now, since we may write:

$$\displaystyle h(x)=\frac{1}{\sqrt{2}}\left(x\sin(x)+x\cos(x) \right)$$

And by the linearity of differentiation, we know:

$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(\frac{d^n}{dx^n}\left(x\sin(x) \right)+\frac{d^n}{dx^n}\left(x\cos(x) \right) \right)$$

Thus, we must have:

$$\displaystyle \frac{d^n}{dx^n}\left(x\cos(x) \right)=x\cos\left(x+n\frac{\pi}{2} \right)+n\sin \left(x+n\frac{\pi}{2} \right)$$


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Home How can I print literal curly-brace characters in python string and also use .format on it?

# How can I print literal curly-brace characters in python string and also use .format on it?

Schitti
1#
Schitti Published in 2011-03-29 00:04:53Z
 x = " \{ Hello \} {0} " print x.format(42)  gives me : Key Error: Hello\\ I want to print the output: {Hello} 42
Greg Hewgill
2#
Greg Hewgill Reply to 2011-03-29 00:08:55Z
 You need to double the {{ and }}: >>> x = " {{ Hello }} {0} " >>> print x.format(42) ' { Hello } 42 '  Here's the relevant part of the Python documentation for format string syntax: Format strings contain “replacement fields” surrounded by curly braces {}. Anything that is not contained in braces is considered literal text, which is copied unchanged to the output. If you need to include a brace character in the literal text, it can be escaped by doubling: {{ and }}.
pajton
3#
 Try this: x = "{{ Hello }} {0}"
Kamil Kisiel
4#
Kamil Kisiel Reply to 2014-03-28 06:34:23Z
 You escape it by doubling the braces. Eg: x = "{{ Hello }} {0}" print x.format(42)
DNR
5#
 Try doing this: x = " {{ Hello }} {0} " print x.format(42)
George Aprilis
6#
George Aprilis Reply to 2015-11-05 11:22:02Z
 Although not any better, just for the reference, you can also do this: >>> x = '{}Hello{} {}' >>> print x.format('{','}',42) {Hello} 42  It can be useful for example when someone wants to print {argument}. It is maybe more readable than '{{{}}}'.format('argument') Note that you omit argument positions (e.g. {} instead of {0}) after Python 2.7
twasbrillig
7#
 The OP wrote this comment: I was trying to format a small JSON for some purposes, like this: '{"all": false, "selected": "{}"}'.format(data) to get something like {"all": false, "selected": "1,2"}  It's pretty common that the "escaping braces" issue comes up when dealing with JSON. I suggest doing this: import json data = "1,2" mydict = {"all": "false", "selected": data} json.dumps(mydict)  It's cleaner than the alternative, which is: '{{"all": false, "selected": "{}"}}'.format(data)  Using the json library is definitely preferable when the JSON string gets more complicated than the example.
tvt173
8#
 If you are going to be doing this a lot, it might be good to define a utility function that will let you use arbitrary brace substitutes instead, like def custom_format(string, brackets, *args, **kwargs): if len(brackets) != 2: raise ValueError('Expected two brackets. Got {}.'.format(len(brackets))) padded = string.replace('{', '{{').replace('}', '}}') substituted = padded.replace(brackets[0], '{').replace(brackets[1], '}') formatted = substituted.format(*args, **kwargs) return formatted >>> custom_format('{{[cmd]} process 1}', brackets='[]', cmd='firefox.exe') '{{firefox.exe} process 1}'  Note that this will work either with brackets being a string of length 2 or an iterable of two strings (for multi-character delimiters).
acacia
9#
 If you have to put braces around a value of a key of the parameter dictionaries or lists of the format method, try the following: >>> "{o}Hello {a}{c}".format(**{"o":"{","c":"}","a":42}) '{Hello 42}'
10#
 Best solution for me is to use this '{ob} Hello {cb} 42'.format(ob='{',cb='}') The reason why because good luck using accepted answer to print something like below. a='Header' ob = '{' cb = '}' str = """ import {a} from '../actions/{a}'; class {a} {ob} constructor() {ob} {cb} somefuncSuccess(){ob} if(true){ob}{cb} {cb} {cb} """ str = str.format(a=a,ob=ob, cb=cb) print(str)  Output: import Header from '../actions/Header'; class Header { constructor() { } somefuncSuccess(){ if(true){} } }
 Reason is , {} is the syntax of .format() so in your case .format() doesn't recognize {Hello} so it threw an error. you can override it by using double curly braces {{}}, x = " {{ Hello }} {0} "  or try %s for text formatting, x = " { Hello } %s" print x%(42)
 Python 3.6 (2017) In the current version of Python one would use f-strings (see also PEP498). Also with f-strings one should use double {{ or }} n = 42 print(f" {{Hello}} {n} ")  produces the desired  {Hello} 42


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# least square method

Keywords: MATLAB Math linear algebra

# least square method

1. What is the essence of least squares
Taking measuring the length with different rulers as an example, several measured values are obtained. How to determine the true value? "y, which minimizes the sum of squares of errors, is the true value." this is based on the fact that if the error is random, the measured value will fluctuate up and down the true value. Ruler example, optimization goal m i n ∑ ( y − y i ) 2 min \sum(y-y_i)^2 min Σ (Y − yi) 2, then the objective function takes the derivative of Y, and y with zero derivative is exactly the arithmetic mean. Least square -- > least + square
2. Generalization: for polynomial fitting
3. MATLAB implementation
3.1 solving equations
Least squares principle and MATLAB code - linear fitting, polynomial fitting and specified function fitting
3.1.1 linear equation
m i n ∑ ( y i − a x i − b ) 2 = 0 min \sum(y_i-ax_i-b)^2=0 min Σ (yi − axi − b)2=0, calculate the partial derivatives of a and B as 0 respectively, and obtain the following equations to solve a and B. In addition, the equations can be written in matrix form.
{ − ∑ 2 ( y i − a x i − b ) x i = 0 − ∑ 2 ( y i − a x i − b ) = 0 \left\{ \begin{array}{c} -\sum{2\left( y_i-ax_i-b \right) x_i=0}\\ -\sum{2\left( y_i-ax_i-b \right) =0} \end{array} \right. {−∑2(yi−axi−b)xi=0−∑2(yi−axi−b)=0 ( ∑ 1 ∑ x i ∑ x i ∑ x i 2 ) ( b a ) = ( ∑ y i ∑ y i x i ) \left( \begin{array}{c} \begin{matrix} \sum{1}& \sum{x_i}\\ \end{matrix}\\ \begin{matrix} \sum{x_i}& \sum{x_{i}^{2}}\\ \end{matrix}\\ \end{array} \right) \left( \begin{array}{c} b\\ a\\ \end{array} \right) =\left( \begin{array}{c} \sum{y_i}\\ \sum{y_ix_i} \end{array} \right) (∑ 1 ∑ xi ∑ xi ∑ xi2) (ba) = (∑ yi ∑ yi xi) 3.1.2 polynomial, which can also be written in matrix form similar to the above. XA=Y, then A=inv(X)*Y
3.2 the optimization problem can be solved by gradient descent method. But the following code does not optimize the step size, which takes time.
clear;
clf;
x0 = linspace(-1,1,100);
% function+Noise
y0 = sin(2*x0).*((x0.^2-1).^3+1/2)+(0.2*rand(1,100)-0.1);
% Raw data image
plot(x0,y0,'.');
% Fitting with polynomial function f(x)
% f(x)=a0+a1*x+...+ak*x^k
% Objective function: sum of squares of errors sum(y-f(x))^2
% Find the optimal value: the partial derivative of each coefficient is 0
% Method 1: solving linear equations XA=Y，be A=Y/X
% Method 2: gradient descent method
% Specify polynomial coefficients k
order = 10;
ini_A = zeros(1,order+1);  % Initial point
% The objective function is to minimize the sum of squares of errors, the initial point is given, and the gradient descent method is used to find the optimal solution
[A,k,err_ols] = GradientMethod_ols(x0, y0, order, ini_A);
y1 = polyfun(x0,order,A);
hold on;
plot(x0,y1);

function [A,k,err_ols] = GradientMethod_ols(x0, y0, order, ini_A)
% x,y: Raw data point
% order,ini_A: A polynomial function is determined
%% Step 1: give the initial iteration point A，Convergence accuracy err，k=1
k = 1;
iterk = 1000;
A = ini_A;
err = 0.5;
%% Step 2:Calculate the gradient and modulus and take the search direction
err_ols = 100;
while(k<iterk)
dir = -grad;  % The search direction is a negative gradient direction
%% Step 3:Make convergence judgment
if err_ols<=err  % The gradient is equal to 0
disp('The error is small enough');
break;
end
%% The fourth step is to find the optimal step size and new iteration points
lmd_opt = 0.1;  % Take the fixed value directly
A0 = A;
while(1)
A = A0 + dir*lmd_opt;
err_ols_1 = sum((y0-polyfun(x0,order,A)).^2);
if err_ols_1<err_ols
err_ols = err_ols_1;
break
end
lmd_opt = lmd_opt*0.5;
end
k = k+1;
end
if k >=iterk
disp('The number of iterations has reached the upper limit!');
end
end

for i = 1:length(A)
end
end

function y = polyfun(x,order,A)
% Value of polynomial function
% y=a0+a1*x+...+ak*x^k
% x independent variable
% k Polynomial order
% A Coefficient vector
y = zeros(1,length(x));
for i =0:order
y = y + A(i+1)*x.^i;
end
end

1. supplement
Line fitting (W=inv(A '* A)*A' * Y / / w = inv (R) * q '* Y) and lsqcurvefit function
Least square method of MATLAB
b.

# Finding the extreme point of multivariate function

I've forgotten that the extreme value of a multivariate function is the partial derivative of each item, so review it.
1. The point where the first derivative is 0 is called the stagnation point, which is not necessarily the extreme point( y = x 3 y=x^3 y=x3), the extreme point is not necessarily the stationary point( y = ∣ x ∣ y=|x| y=∣x∣)； For multivariate functions, stationary points are all points where the first derivative is zero.
2. The point where the curve convexity changes is the inflection point. If the function is second-order continuous differentiable, the inflection point can deduce that the second-order derivative is 0;
3. The extreme point is the abscissa of the upper maximum or minimum point in a certain sub interval of the function image, which appears at the stationary point or non differentiable point.
Extremum of multivariate function and its solution
( f ( x ) = x 2 , f ′ ′ ( x ) 2 = 4 > 0 , f ′ ′ ( x ) = 2 > 0 have extremely Small value F (x) = x ^ 2, f '' (x) ^ 2 = 4 > 0, f '' (x) = 2 > 0 has a minimum f(x)=x2,f ′′′ (x) 2 = 4 > 0, f ′′ (x) = 2 > 0 (with minimum)
If the known function has and has only one extreme point, then "find the point with partial derivative of 0 to get the extreme point", is that right?

Posted by britey on Fri, 19 Nov 2021 02:12:36 -0800