Equation of plane question?

Distance between a point and a plane?

• Let W be the plane through the origin in R3 spanned by the set S = {(1, 1, 1), (1, 0, -1)} Find the distance from the point P=(-1, 1, -2) to the plane W I calculated this by using cross product to find the plane equation of -x+2y-z=0. Putting this into the distance equation |a+b+c+d|/sqrt(a^2+b^2+c^2) I got an answer of 4/sqrt(6) ~ 1.63 I'm confused for two reasons. According to the answer key I'm wrong, it should be sqrt(150)/6 ~ 2.04. If I had gotten an answer of 5/sqrt(6) it would have given the same answer which makes it even weirder. Then the next question asks for the equation of the plane meaning that there is some other way I should be doing this. The equation I got earlier was the answer given. I'm not sure if this relevant but the question before that asked for the projection of the point onto the set which was -1/6 (1, 4, 7)

• Answer:

  I'm a bit rusty on this, but isn't the formula for the distance of (x₁, y₁, z₁) from ax + by + cz -d = 0 Dist = |ax₁ + by₁ + cz₁ - d|/√(a² + b² + c²) ? With a = -1, b = 2, c = -1, d = 0 and the point (-1, 1, -2) that gives (1 + 2 + 2)/√6 = 5/√6 which, as you have observed already, is equal to (√150)/6 PS. If you apply the Pythagoras distance formula to the points -1/6 (1, 4, 7), (-1, 1, -2) you get (√150)/6, thus verifying that the distance of a point from a plane is its distance from its projection onto the plane. So maybe you were meant to use this method to find the distance, and then after that find the equation of the plane. But your method is valid, of course.