Vectors and the Geometry of Space

Question

Find the distance from the point (-1, -4, 0) to the plane −3x −y + 3z =3

Answer 1

Let create normal equation of plane: D = √9 + 1 + 9 = √19. We dividing by D: (- 3x - y + 3z - 3)/√19 = 0.

Now distance d = I- 3·(- 1) - (- 4) + 3·0 - 3I/√19 = 4/√19

Answer 2

Find the distance from the point (-1, -4, 0) to the plane −3x −y + 3z =3

SOLUTION

The idea behind this solution is that by selecting a second point on the plane , say Q (0 , 0 ,1 ) ,and

if we call the given point P ( -1, -4, 0 ).then the distance of the point P to the given plane , will be equal

to the length of the projection of the vector PQ on the normal vector n = < -3, -1, 3 > of the plane.

PQ =< 1, 4, 1>

d = || proj_n PQ || = | PQ • n | / || n|| = | -3 -4 +3 | / √19 = 4 / √19