Byron S. answered • 10/29/14
Math and Science Tutor with an Engineering Background
The cross product of two vectors produces a vector that is perpendicular to both of the original vectors. In this problem, you can find two vectors in the plane using the points given. Then you can take the cross product to find a vector orthogonal to that plane, and normalize it to make it a unit vector.
Given two points, you can find the vector that points from one to the other by subtracting their components. Which pairs of points you use does not matter, nor does the order matter in this case. All 6 possible vectors by subtracting pairs of these points will lie in the desired plane.
v = P - Q = ⟨-5 - -1, -5 - -1, -4 - 0⟩ = ⟨-4, -4, -4⟩
w = Q - R = ⟨-1 - -1, -1 - -1, 0 - -5⟩ = ⟨0, 0, 5⟩
Cross products are a special type of determinant which would be difficult to type into this text box. You can look up the formula in your textbook or online.
a = v × w = ⟨(-4)*5+(-4)*0, (-4)*5+(-4)*0, (-4)*0+(-4)*0⟩
a = ⟨-20, -20, 0⟩
|a| = √((-20)^{2} + (-20)^{2} + 0^{2}) = √(800) = 20√(2)
u = a / |a| = ⟨-1/√(2), -1/√(2), 0⟩
The problem asks for the vector with a positive first coordinate, so multiply through by -1. This is in the opposite direction, which is orthogonal to the plane, on the other side.
u = ⟨1/√(2), 1/√(2), 0⟩