# Plane from intersection line and point

The task: Determine the plane containing point $P( -5 , 2 , 3 )$ and going through the intersection line of the planes $2x + y + 5z = 31$ and $-4x + 5y + 4z = 50$

1.: Intersect the two given planes, resulting in a line in parameter form ( $X = P + t * V$ )

2.: Determine two arbitary points on the line

3.: Form the new plane using the two points on the line and the given point P

(by creating an equation system $ax + by + cz + d = 0$, with $3$ equations with the $x$, $y$ and $z$ values of the points inserted)

• Is my approach for solving the problem right? (I don't ask for a solution!)
-
Yes, your approach is pretty much right. From the two points on the line and the point $P$, you can then find two vectors ("starting" at the same point) and take the cross product of those two to get a normal vector for the plane. –  Thomas Apr 18 '12 at 18:41
The procedure will work. It involves somewhat more effort than necessary. –  André Nicolas Apr 18 '12 at 18:44

(1) You find the intersection of the two planes and find (say parametric) equations for the line of intersection.

(2) You find two (distinct) points on the line call them $A$ and $B$.

(3) Then you can find a normal vectors for the plane that you are seeking by finding $\stackrel{\to}{AB}$ and $\stackrel{\to}{AP}$ and the $\vec{n} = \stackrel{\to}{AB}\times \stackrel{\to}{AP}$.

But you could also just use that the cross product of the two normal vectors (say $\vec{n}_1$ and $\vec{n}_2$) for the two given planes is "contained" in the plane. So you really just need one point (say $A$) on the line of intersection and then this vector. So a normal vector would be $(\vec{n}_1\times \vec{n}_2)\times \stackrel{\to}{AP}$

-
For any two parameters $\lambda, \mu \in \mathbb{R}$, not both zero, the combination
$$\lambda \cdot (2x+y+5z) + \mu\cdot(−4x+5y+4z)= \lambda \cdot 31 + \mu \cdot 50$$
is the equation of a plane that has the same intersection line as the two given planes. (This is called the pencil of planes through that line.) Substitution of the point $(−5,2,3)$ gives:
$$\lambda \cdot 7 + \mu \cdot 42 = \lambda \cdot 31 + \mu \cdot 50$$
or $24 \lambda + 8 \mu = 0$. So you can take $\lambda = 1$ and $\mu=-3$ to find the requested plane.