Question

Given the plane 2x-3y+7z=4,

a) Find the equation for the line perpendicular to the plane that intersects the plane at the point P=(6,5,1).
b) How far is it from the point (4,-2,3) to the plane?

Answer 1

a)(x-6)/2=(y-5)/(-3)=(z-1)/7
b)The normal equation of the plane
(2/sqrt(62))x-(3/sqrt(62))y+(7/sqrt(62))z-
-4/sqrt(62)=0 62=2^2+(-3)^2+7^2

The distance=(2*4-3*(-2)+7*3-4)/sqrt(62)=
sqrt(31/2)

Answer 2

why do you have = signs in part a?