dannyboi2006tk

2022-10-08

Let L be the line of intersection between the planes x + 3y = 7 and 2y + z = 4, and let P be the point (3, 1, 1).
Determine the equation (in normal form) of the plane containing L and P.
I struggle with getting vectors from the two planes to multiply, can anyone show me how?
$L=\left(1,2,0\right)+t\left(3,-1,2\right)$. But i somehow fail to see how my book got that.

### Answer & Explanation

Emmanuel Russo

To obtain L, parameterize it with $y=2+t$ and, from the two given planes, we have $x=1-3t$ and $z=-2t$, i.e.
$L=\left(1,2,0\right)+t\left(-3,1,-2\right)$
with $Q\left(1,2,0\right)$ and $\stackrel{\to }{d}=\left(-3,1,-2\right)$. Let $A\left(x,y,z\right)$ be any point on the plane and $P\left(3,1,1\right)$. Then, the normal vector of plane is
$\stackrel{\to }{n}=\stackrel{\to }{QA}×\stackrel{\to }{QP}=\left(x-1,y-2,z\right)×\left(2,-1,1\right)\phantom{\rule{0ex}{0ex}}=\left(-2+y+z,1-x+2z,5-x-2y\right)$
which satisfies $\stackrel{\to }{n}\cdot \stackrel{\to }{d}=0$, i.e.
$\left(-2+y+z,1-x+2z,5-x-2y\right)\cdot \left(-3,1,-2\right)=0$
which yields the equation of the plane
$x+y-z=3$

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