Find the scalar equation of the plane containing the point (6,1,-5) and that is perpendicular to the line...?

Question

passing through the points P(-2,4,3) and Q(3,3,7)..

Answer 1

The directional vector of the line thru the two points P(-2,4,3) and Q(3,3,7) will also be the directional vector of the plane perpendicular to it.

Let n be the directional vector of the line and normal vector of the plane. Then

n = PQ = <3-(-2), 3-4, 7-3> = <5, -1, 4>

The equation of the plane is:

5(x - 6) - 1(y - 1) + 4(z - (-5)) = 0
5x - 30 - y + 1 + 4z + 20 = 0
5x - y + 4z - 9 = 0

Answer 2

Dude, is that your homework?
If the line passes through P and Q, it is parallell to the vector P-Q=(-5,1,-4), right? So, the plane perpendicular to this vector should have an eq. like -5x+y-4z=C. Use the point that's given to find the value for C...