# 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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Take the cross product of the normal vector given and the vector you get from these two points that lie on the plane. This will give you the normal vector for the plane you want to find the equation for.

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