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Let's say you have two intersecting planes. I know that to find the line intersection of them is simply the cross product of their two gradients.

Let's say we get the Line $L=(2,3,0)+t<1, -4 ,\ 2>$.

Now we want to find a plane that contains the line L and is perpendicular to x+2y-3z=1.

Is my work correct?: I do $<1,-4,2>\times <1,2,-3> = <a,b,c> $ and then parametrize the plane as $P(t,s)=(2,3,0)+t<1, -4 ,\ 2> + s<a,b,c>$.

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Eliot Behr
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1 Answers1

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Let $\vec{n} = (1,2,-3)$ and $\vec{a} = (1,-4,2)$. Then, $\vec{a}\times\vec{n}=(8,5,6)$ is the vector normal to the plane that is parallel to the line $L$ and perpendicular to $x+2y-3z=1$ is .

Denote the point in the line $P_0(2,3,0)$. For any point $P(x,y,z)$ in this plane, the vector $\vec{PP_0}$ has to be perpendicular to $\vec{a}\times\vec{n}$, i.e.

$$ (\vec{a}\times\vec{n})\cdot \vec{PP_0} = 0 $$

Thus, the equation of the plane is given by,

$$(8,5,6)\cdot (x-2,y-3,z) = 0$$

or, in the standard form $8x+5y+6z=31$.

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