Question

Find an equation of the plane orthogonal to the line:
(x,y,z) = (5,-6,-1) + t(3,10,-3) which passes through the point (-6,-1,5).

so i got so far:
x=5+3t
y=-6+10t
z=-1-3t

Should be in form ax+by+cz+d=d
Not sure where to go from there...

Answer 1

If the plane is orthogonal to
(5,-6,-1) + t(3,10,-3), then it is also orthogonal to the parallel line
t(3,10,-3) which moves through the origin. Consider first the plane orthogonal to this line which contains the origin. If the point (x,y,z) is on that plane, then:

(x,y,z) dot (3,10,-3) = 0 ---->

3x + 10 y - 3z = 0

The plane we are seeking is parallel to this plane and is therefore given by the equation:

3x + 10 y - 3z = d

You can compute d by demanding that the point (-6,-1,5) lies on the plane.