Question

How do you find parametric equations for the line of intersection of two planes $2x-2y+z=1$ and $2x+y-3z=3$ ?

Answer 1

We have been given the equations of two planes whose parametric equations for the line of intersection is to be found. Firstly, we shall find the point of intersection of these planes setting the value of the variable-z equal to zero. Then we shall find the cross product of the normal vectors of the planes to find the direction vector for the line. Further, we shall form the parametric equations of this line of intersection.

Complete step-by-step answer:
Given two planes, $2x-2y+z=1$ and $2x+y-3z=3$.
The parametric equations of their line of intersection is given as
$\begin{aligned} & x={{x}_{0}}+\lambda a \\\ & y={{y}_{0}}+\lambda b \\\ & z={{z}_{0}}+\lambda c \\\ \end{aligned}$
Where,
$\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)=$ point of intersection of the two planes
$\left( a,b,c \right)=$ direction vectors of the line of intersection
$\lambda =$ parameter of the parametric equations
In order to find the point of intersection of the two planes, we shall set $z=0$ in both equations and get the modified equations as $2x-2y+\left( 0 \right)=1$ and $2x+y-3\left( 0 \right)=3$, that is,$2x-2y=1$ and $2x+y=3$.
Now, we shall simply solve these equations simultaneously to find the x and y-coordinate of the point of intersection of planes.
$\begin{aligned} & \text{ }2x-2y=1 \\\ & \underline{-2x+\text{ }y=3} \\\ & \text{ 0 }-3y=-2 \\\ \end{aligned}$
$\Rightarrow -3y=-2$
$\Rightarrow y=\dfrac{2}{3}$
Substituting this value of y-coordinate in one of the equations, we get
$\begin{aligned} & \Rightarrow 2x-2\left( \dfrac{2}{3} \right)=1 \\\ & \Rightarrow 2x-\dfrac{4}{3}=1 \\\ & \Rightarrow 2x=1+\dfrac{4}{3} \\\ \end{aligned}$
$\Rightarrow 2x=\dfrac{7}{3}$
Dividing both sides by 2, we get
$\Rightarrow x=\dfrac{7}{6}$
Thus, the point of intersection is $\left( \dfrac{7}{6},\dfrac{2}{3},0 \right)$ .
Now to find the direction vector of line, we shall find the cross product of normal vectors of $2x-2y+z=1$ and $2x+y-3z=3$.
$\Rightarrow \left( a,b,c \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 2 & -2 & 1 \\\ 2 & 1 & -3 \\\ \end{matrix} \right|$
Expanding along the first column, we calculate the determinant as
$\Rightarrow \left( a,b,c \right)=\hat{i}\left( -2\left( -3 \right)-1\left( 1 \right) \right)-\hat{j}\left( -3\times 2-2\times 1 \right)+\hat{k}\left( 2\times 1-2\left( -2 \right) \right)$
$\Rightarrow \left( a,b,c \right)=\hat{i}\left( 6-1 \right)-\hat{j}\left( -6-2 \right)+\hat{k}\left( 2+4 \right)$
$\Rightarrow \left( a,b,c \right)=5\hat{i}+8\hat{j}+6\hat{k}$
Finally, we shall substitute these values in the parametric equations.
$\begin{aligned} & x=\dfrac{7}{6}+5\lambda \\\ & y=\dfrac{2}{3}+8\lambda \\\ & z=0+6\lambda \\\ \end{aligned}$
To find the combined equation, we shall add these three separate equations.
$r=\dfrac{7}{6}+\dfrac{2}{3}+\lambda \left( 5\hat{i}+8\hat{j}+6\hat{k} \right)$
Therefore, parametric equations for the line of intersection of two planes $2x-2y+z=1$ and $2x+y-3z=3$ are $x=\dfrac{7}{6}+5\lambda ,y=\dfrac{2}{3}+8\lambda$ and $z=6\lambda$ or $r=\dfrac{7}{6}+\dfrac{2}{3}+\lambda \left( 5\hat{i}+8\hat{j}+6\hat{k} \right)$.

Note: We must remember to set the value of one variable out of the three variables, x, y and z, equal to zero while computing the point of intersection of two planes or else it will be impossibly to solve the two equations. However, another method of finding the point of intersection is by application of the Cramer’s rule.