# Find an equation of the plane passing through 2 points and perpendicular to another plane

Find an equation of the plane that passes through the points $(0-2,5)$ and $(-1,3,1)$ and is perpendicular to the plane $2z = 5x + 4y$.

Here's what I have so far:

The plane through $(0,-2,5)$ is $ax + b(y+z) + c(z-5) = 0$. And the plane also passes through $(-1,3,1)$ so I get: $$-a + 5b - 4c = 0 \tag{1}.$$

When I looked at the explanation it says:

Now we know that the plane is perpendicular to $5x + 4y - 2z = 0$ and then it replaces $(x,y,z)$ with $(a,b,c)$ to get $$5a + 4b - 2c = 0. \tag{2}$$

It continues from there saying to solve the two equations to get $\frac{a}{6} = \frac{b}{-22} = \frac{c}{-29}$.

I know how to solve it once it gets to this but I have absolutely no idea how they got to this step.

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possible duplicate of this –  chaohuang Sep 20 '12 at 15:10

The explanation is using these three facts:

1. Two planes are perpendicular iff their normal vectors are perpendicular.
2. If the equation of a plane is $ax+by+cz=d$, then a normal vector to the plane is $(a,b,c)$.
3. Two vectors $(a,b,c)$ and $(d,e,f)$ are perpendicular iff $ad+be+cf=(a,b,c)\cdot(d,e,f)=0$.

So when the book asserts that the plane given by $ax + b(y+2) + c(z-5) = 0$ is perpendicular to the plane $5x + 4y - 2z = 0$, it does so with the equation $5a+4b-2c=0$.

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First take a direction vector to those two points, then, take the given normal from the equation of a plane. The cross product of the direction vector with the normal will then gives the normal of the desired plane, finally take either of the two point and form a point normal equation of a plane.

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how to solve the eqtn ?

$5a + 4b -2c =0$

$-a +5b - 4c =0$

take these eqtns and make "a" terms alike

like this

$-5a - 4b +2c =0$

$-5a +25b -20c =0$

from this we get $-5a/(-4*-20)-(2*25)$ which we will get as $-5a/30$

similarly make b terms alike and you will get $20b/440$

similary

finally it will giv u

$(a/6) = (b/-22) = (c/-29)$

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