Find a parametric representation for the line of intersection of the planes with equations $x+y+z = 3$ and $2x-y-z = 5$, and find the point where the line intersects the plane with equation $x-y = 2$.
First I get the two norm vectors of the given planes, $v = (1,1,1)$ and $u = (2,-1,-1).
$$\begin{align*}
u\times v &= \begin{pmatrix} i&j&k \\ 1&1&1 \\ 2&-1&-2 \end{pmatrix} = \begin{pmatrix} 1&1 \\ -1&-2\end{pmatrix}i – \begin{pmatrix}1&1 \\ 2&-2 \end{pmatrix}j + \begin{pmatrix}1&1 \\ 2& -1 \end{pmatrix}k \\
&= -1i + 4j – 3k \\
&= (-1,4,-3)
\end{align*}$$
Now after getting the cross product, I have the direction of the line of intersection. Now I need to find the point where the line intersects the plane, so taking the norm vector of $x – y = 2$ gives $(1,-1,0)$ so this can serve as my point of intersection?
Then the parametric representation is $= t(-1,4,-3) + (1,-1,0)$. Did I use all the information correctly or am I understanding parts of this problem wrong?
Best Answer
We are working with \begin{cases} \pi_1: & x+y+z=3\\ \pi_2: & 2x-y-z=5\\ \end{cases}
Since their normal vectors are non-parallel, we know that the planes are not parallel. As such, we know they intersect in a line at some point with some direction vector.
Suppose the direction vector can be written as some scalar multiple of a vector whose $z$ component is $1$. So, we can let $\boxed{z=t}$. Substituting this into the two equations we get
\begin{align} x+y+t&=3\\ 2x-y-t&=5\\ \end{align}
With these two equations, we can solve for $x$ and $y$ in terms of $t$, and those will become the parametric equations of our line. I will leave the solution as an exercise (use either the substitution or elimination method to solve the linear system in terms of $t$). The equations we end up with are $\boxed{x=\frac83}$ and $\boxed{y=\frac13-t}$
These three boxed equations are the parametric equations for the line of intersection. $$l:\begin{cases} x=\frac83\\ y=\frac13-t\\ z=t\end{cases} $$
Since we want the point of intersection between $l$ and the given plane $x-y=2$, simply substitute the parametric equations of $l$ in and solve for $t$, and then finally substitute that $t$ value back into the parametric equations.
\begin{align} x-y&=2\\ (\frac83)-(\frac13-t)&=2\\ \frac73+t&=2\\ t&=-\frac13 \end{align}
Therefore, we have the point $(x,y,z)=(\dfrac83,\dfrac13-t,t)=\boxed{(\frac83,\dfrac23,-\frac13)}$.
To check: we know that the point $(\frac83,\frac23,-\frac13)$ has to be a point that exists on all three planes. As we expect, it satisfies all three equations. This hints that we could have solved the planar system without first solving for the line of intersection of the first two planes!
Simply put, we want to find the solution for the system of linear equations \begin{cases} x+y+z&=3\\ 2x-y-z&=5\\ x-y&=2 \end{cases}