Question

Find an equation of a plane through the point (4, -4, 1) which is orthogonal to the line x=-4+5t, y=4-3t, z=-2+4t, in which the coefficient of x is 5.

Answer 1

I'm not sure of the orthodox way to do this, but I think I got the solution...

Answer 2

ok

Answer 3

the vector that points in the line given is (5,-3,4), agreed?

Answer 4

yes i agree

Answer 5

sorry wait I have to rethink :S

Answer 6

its ok

Answer 7

okay...since that vector is to be perpendicular to the plane, we can call it our normal vector n=(5,-3,4)

now we need a vector in the plane. the position vector of the given point is Po=(4,-4,1)
and we can call any other point in the plane P=(x,y,z)
the vector P-Po will therefor be in the plane, which means that
n(dot)(P-Po)=0

Answer 8

yes

Answer 9

that translates to
(5,-3,4)(dot)(x-4,y+4,z-1)=0
5(x-4)-3(y+4)+4(z-1)=0
5x-20-3y-12+4z-4=0
5x-3y+4z=36

I think I did that right, but it has been a while
if it makes sense to you though it's probably good... I hope ;)

Answer 10

yes it makes sense thank you

Answer 11

Its correct thank you :)

Answer 12

awesome :) very welcome, thanks for the practice!