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