Find equation of a plane that passes through point P $(-1,4,2)$ that contains the intersection line of the planes
$$\begin{align*}
4x-y+z-2&=0\\
2x+y-2z-3&=0
\end{align*}$$
Attempt:
I found the the direction vector of the intersection line by taking the cross product of vectors normal to the known planes. I got $\langle 1,10,6\rangle$. Now, I need to find a vector normal to the plane I am looking for. To do that I need one more point on that plane. So how do I proceed?
Best Answer
Put the two given equations together in a system, set $x=0$ (say) and solve for $y,z$ to get an actual point on the intersection line. Accompanied by its direction vector and the outlying point $P$, we can determine three points from the desired plane and subsequently determine an equation for it.
Dostre's edit (see comments)
$x=0:\begin{cases} y-2z-3=0 \\ -y+z-2=0 \end{cases}$ ;$\;\;\;\;$ add them up and you get:
$\;\;\;\;\;\;\;\;\;\;\begin{cases} -z-5,\;\;z=-5\\ -y+z-2=0,\;y=-7 \end{cases}\Rightarrow$Point $ (0,-7,-5)$, call it Q, is on the line of intersection.
So now we have two points $P(-1,4,2)$ and $Q(0,-7,-5)$ on our desired plane and vector w thats is parallel to the desired plane. In order to find the equation of the desired plane we need a vector that is normal to it.We can find that normal vector by taking cross product of two vectors that are parallel to the desired plane. We already have w so the other vector will be
*PQ*$<0-(-1),-7-4,-5-2>=<1,-11,-7>$
Now normal vector to desired plane will be the cross product of w and PQ:
PQ x w=$\begin{vmatrix} i & j & k \\ 1 & -11 & -7 \\ 1 & 10 & 6 \end{vmatrix}=i\begin{vmatrix} -11 & -7 \\ 10 & 6 \end{vmatrix}-j\begin{vmatrix} 1 & -7 \\ 1 & 6 \end{vmatrix}+k\begin{vmatrix} 1 & -11\\ 1 & 10 \end{vmatrix}=4i-13j+21k$
$4i-13j+21k=<4,-13,21>$ is the vector normal to the desired plane.