Find value of c; Plane perpendicular to the line

Question:

Find the value of c so the plane perpendicular to the line x =3+t y=4t z=-4-t that contains the point (1,5,6) also contains the point (c,3c,5c).

Attempt at solution:

First I found the vector of the line:

Point P when t =0 (3,0,-4)
Point Q when t=1 (4,4,-5)

Vector PQ <4-3,4-0,-5-(-4)> --> <1,4,-1>

So to get the plane through the first point I just used the vector and the point to get => 1(x-1)+4(y-5)-1(z-6) = 0

Then I used the same vector and the second point to get =>

1(x-c)+4(y-3c)-1(z-5c) = 0

This is where I got stuck. I set the two equations equal to each other:


1(x-1)+4(y-5)-1(z-6) =1(x-c)+4(y-3c)-1(z-5c)


...but in order to solve for c I need to know what x, y, and z are supposed to be. I would substitute the parametric equations for x, y, and z, but that would only be if I was solving for t, right?

Any suggestions? Thanks

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**Update: I realized I could cancel out all of the x,y, and z terms to give me c =15/8 can anyone tell me if I took the right approach to solving this problem? Thanks

I meant to put 15/8

Quote:

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Originally Posted by CocoLily

Question:

Find the value of c so the plane perpendicular to the line x =3+t y=4t z=-4-t that contains the point (1,5,6) also contains the point (c,3c,5c).

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The equation of that plane is . WHY?

Quote:

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Originally Posted by Plato

The equation of that plane is . WHY?

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Equation of plane when

vector = <a,b,c> = <1,4,-1> and point = (x0,y0,z0) = (1,5,6)

Quote:

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Originally Posted by CocoLily

Equation of plane when

vector = <a,b,c> = <1,4,-1> and point = (x0,y0,z0) = (1,5,6)

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So solve for .
.

Quote:

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Originally Posted by Plato

So solve for .
.

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Ah okay cool, I think I did it right. Your approach is a lot easier though, haha!