Well QO = -OQ = (0, 0, 1). If you want to write this as the sum of two vectors (one of which you know, n, and one of which you don't, a), you have n + a = -OQ. Surely you can figure out what a has to be.
Hi guys, this is part c of a series of problems, where I'm now stuck.
Let P be the plane 2x + y - z = 1
Point Q is at (0, 0, -1)
Take n = 2i + j - k as a normal vector of the plane P.
Decompose the vector QO into the sum of two vectors; one of them is parallel to n and the other one is orthogonal to n.
Any help greatly appreciated!
Given two non-zero, non-parallel vectors it possible to do such a decomposition:
$\overrightarrow {{B_\parallel }} = \dfrac{{\overrightarrow A \cdot \overrightarrow B }}{{\overrightarrow B \cdot \overrightarrow B }}\overrightarrow B \,\& \,\overrightarrow {{B_ \bot }} = \overrightarrow A - \overrightarrow {{B_\parallel }} $
is the answer 1/6(2i+j-k) for parallel and -1/6(2i+j+5k) for orthogonal?. If so, do we get the 2i+j+5k with trial and error? how would we explain in our answer how we got 2i+j+5k?
Thanks