Vectors-finding angle and distance from a point

Question

Use the following information to answer questions (1) and (2).

Plane A passes through points (1, 0, 0),(0, 1, 0),(0, 0, 1)

Plane B is perpendicular to the vector n = 2i + j − 2z.

(1) What is the angle (in radian) between Plane A and Plane B?

(2) What is the distance from the point (0, 1, 2) to Plane A?

Answer 1

Once again, a glance at the points tells us the equation of the plane: x + y + z = 1 (a very well-known plane!)

Then, (1, 1, 1) . (2, 1, -2) / √((1^2+1^2+1^2)*(2^2+1^2+(-2)^2)) = 1/(3√3) = √3/9

Then, the angle is cos^-1(√3/9) = 78.90°

x+y+z for (0, 1, 2) = 3 Recall that A is defined by x+y+z = 1

Then, the distance from the point to the line is (3 - 1)/√(1^2+1^2+1^2) = 2/√3 = 2√3/3