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!)
  • 2
    $\begingroup$ 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. $\endgroup$
    – Thomas
    Commented Apr 18, 2012 at 18:41
  • $\begingroup$ The procedure will work. It involves somewhat more effort than necessary. $\endgroup$ Commented Apr 18, 2012 at 18:44

2 Answers 2


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.


To write a formal answer to your question.

Your approach sounds right. So

(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 $(\vec{n}_1\times \vec{n}_2)\times \stackrel{\to}{AP}$


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