# Thread:

1. ## easy way of finding line of intersection of 2 planes

I may be confused so I am hoping someone could verify what I am saying is correct.

To find the line of intersection of 2 planes subtract one from the other so that one of the variables cancels out. Then introduce a parameter and then solve for x y and z.

e.g.
$x+2y-z=10$ and $2x+3y+z=0$. Subtracting 2 times the first equation from the second gives $-y+3z=-20$. Let z = t. Solving the equation for y give y=20+3t. Then from the first equation x=10-2(20+3t)+t
am I correct in believing that I have found the parametric equation of the line of intersection of the 2 plains mentioned above?

The reason I'm doubtful is because when I google searched "line of intersection of two planes" I found a more difficult approach that $r=r_o+tv$ where the cross product is used to find v. I guess this way is used to find the vector equation of the line?

2. Hello, superdude!

To find the line of intersection of 2 planes, subtract one from the other,
so that one of the variables cancels out.
Then introduce a parameter and then solve for x y and z.

.Example: . $\begin{array}{cccc}x+2y-z&=&10 & [1]\\<br /> 2x+3y+z&=&0 & [2]\end{array}$
Let me show you the way I explain it . . .

$\begin{array}{cccccc}\text{Multiply }2\times [1]\!: & 2x + 4y - 2z &=& 20 \\ \text{Subtract [2]:} & 2x + 3y + z &=& 0 \\ \\[-3mm]<br /> \text{And we have:} & \qquad\; y - 3z &=& 20 \end{array}$

. . Hence: . $y \:=\:3z + 20$

Substitute into [1]: . $x + 2(3z+20) - z \:=\:10$

. . Hence: . $x \:=\:\text{-}5z-30$

So we have: . $\begin{array}{ccc}x &=& \text{-}5z - 30 \\ y &=& 3x + 20 \\ z &=& z \end{array}$

On the right, replace $z$ with a parameter $t.$

. . and we have: . $\begin{Bmatrix}x &=& \text{-}5t - 30 \\ y &=& 3t + 20 \\ z &=& t \end{Bmatrix}$

The reason I'm doubtful is because when I google searched
"line of intersection of two planes" I found a more difficult approach
where the cross product is used to find v. I guess this way is used
to find the vector equation of the line?
Yes, but you may find this vector approach is faster.

The normal vectors of your two planes are: . $\langle1,2,-1\rangle\,\text{ and }\,\langle2,3,1\rangle$

The cross-product gives the direction of the line of itersection.

. . $\left|\begin{array}{ccc}i & j & k \\ 1 & 2 & \text{-}1 \\ 2 & 3 & 1 \end{array}\right| \;=\;5i - 3j - k \;=\;\langle 5,\text{-}3,\text{-}1\rangle$

Now find any point that lies on both planes . . . For example: . $(0,2,\text{-}6)$

And we have our parametric equations: . $\begin{Bmatrix}x &=& 0 + 5t \\ y &=& 2 -3t \\ z &=& \text{-}6 - t \end{Bmatrix}$

3. Originally Posted by Soroban

Now find any point that lies on both planes . . . For example: . $(0,2,\text{-}6)$
How did you find a point so quickly?
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