In order to solve problems involving vectors and their resultants and dot products, it is necessary to
A typical problem involving vectors which can be solved by adding and/or subtracting vectors, using a dot product, or applying the Law of Sines or Law of Cosines to give us information about vectors that form a non-right triangle. Usually you are asked to find information about unmeasured components and angles. Note that we will represent vectors either by using bold type v
or by using the vector notation
Let's look at two introductory examples of this type of problem.
First we make a diagram in standard position. The attempted flight path of the plane
is vector OX
. Notice that the angle
shown is 50° because the 40° bearing is measured clockwise from North. The vector XY
represents the wind from the South. The vector OY
is the resultant path of the plane, affected by the wind. In the diagram shown below, the known magnitude and angle
are labeled. It is useful to put such a diagram in “standard position” with North being at the top of the vertical axis. We then measure a bearing clockwise from North to label the measurement of an angle.
We need to find vector OY
(ground speed) which is the vector
sum of OX
(air speed) and XY
(wind speed). One way to do this is to add the components of OX
and find the magnitude and direction of OY
, their resultant..We have the following :
given above could also have been written as:
We find the magnitude of OY, the plane’s ground speed, using the Pythagorean Theorem:
makes with the horizontal axis
in the diagram can be found as follows:
has been pushed (54.194° - 50º) = 4.194º off its attempted path if no corrections are made for the wind. The plane's new bearing is
90° – 54.194° = 35.806°
The final bearing and ground speed of the plane are 35.806° at 439.479 mph.
We first make a vector
diagram in standard position.
made by vector OY
with the horizontal axis
is 50° because bearings are measured clockwise from North. The new bearing is the direction the plane
should take in order to allow the wind to push it upward to the desired 40° bearing, which is the angle
made by vector OY
with the vertical (North) axis.
We do not know the angle
formed by the vector OX
and the horizontal axis, labeled as (50º - θ) in our diagram.
In our diagram, we labeled angle
YOX as angle
θ. Using triangle
YOX, we can apply the Law of Sines as follows:
must therefore aim 4.6° East of the desired 40° bearing, or at a new bearing of
(40º + θ) = 44.6°
To find the magnitude of vector OY
, we must first know the measure of the angle
in the triangle
OXY. To do this, we will start with the fact that the sum of the angles in triangle
equals 180º. Note that angle b
= 40º due to vertical angles
ÐOXY = 180° - b - θ = 180 - 40° - 4.6° = 135.4°
We will now use the Law of Sines once again to find the magnitude of the vector OY
Remember that when working vector problems that involve currents the vector equation is:
air speed + wind speed = ground speed
OX + XY = OY
In order to reach its desired destination, our plane will fly with
an air speed of 400 mph on a bearing of 44.6º.
His resultant ground speed will be 436.9 mph at a bearing of 40º.
Now that we can determine the components of the vectors OY
we can use their dot product
to check our calculations and verify that the angle
YOX (θ in our diagrams) is 4.6º.
OY = (436.9cos(50°), 436.9sin(50°)) = (280.8, 334.7)
OX = (400cos(45.4°), 400sin(45.4°)) = (280.9, 284.8)
Recall that in general, if v
) and u
), are two vectors, their dot product
can be expressed as either
represent the magnitudes
of vectors v
Applying the second formula to our situation results in the formula
Since this is a cumbersome calculation, you should use a calculator. Entering this into a calculator is shown in the screen below.
The result is cos(θ)
0.9967788785. This gives θ
In some situations, the Laws of Sines and Cosines work very well. In other situations, it might be necessary to first calculate the vector