# Find value of c; Plane perpendicular to the line

• Nov 14th 2009, 11:53 AM
CocoLily
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
• Nov 14th 2009, 11:58 AM
Plato
Quote:

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).

The equation of that plane is $(x-1)+4(y-5)-(z-6)=0$. WHY?
• Nov 14th 2009, 01:01 PM
CocoLily
Quote:

Originally Posted by Plato
The equation of that plane is $(x-1)+4(y-5)-(z-6)=0$. WHY?

Equation of plane $a(x-x0)+b(y-y0)+c(z-z0)=0$ when

vector = <a,b,c> = <1,4,-1> and point = (x0,y0,z0) = (1,5,6)
• Nov 14th 2009, 01:10 PM
Plato
Quote:

Originally Posted by CocoLily
Equation of plane $a(x-x0)+b(y-y0)+c(z-z0)=0$ when

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

So solve for $c$.
$(c-1)+4(3c-5)-(5c-6)=0$.
• Nov 14th 2009, 01:27 PM
CocoLily
Quote:

Originally Posted by Plato
So solve for $c$.
$(c-1)+4(3c-5)-(5c-6)=0$.

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