Vector projections

1. Feb 13, 2008

tony873004

Let $$\overrightarrow v$$ and $$\overrightarrow w$$ be vectors in R3. Prove that $$\overrightarrow w - {\rm{proj}}_{\overrightarrow v } \overrightarrow w$$ is perpendicular to $$\overrightarrow v$$ .

Here's my attempt:
$$\begin{array}{l} \left( {\overrightarrow w - {\rm{proj}}_{\overrightarrow v } \overrightarrow w } \right) \cdot \overrightarrow v \mathop = \limits^? 0 \\ \\ {\rm{proj}}_{\overrightarrow v } \overrightarrow w = \frac{{\overrightarrow v \cdot \overrightarrow w }}{{\left| {\overrightarrow v } \right|^2 }}\overrightarrow v \\ \\ \overrightarrow v \cdot \overrightarrow w = v_1 w_1 + v_2 w_2 + v_3 w_3 \\ \\ \left| {\overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + v_3^2 } \\ \left| {\overrightarrow v } \right|^2 = v_1^2 + v_2^2 + v_3^2 \\ \\ \frac{{\overrightarrow v \cdot \overrightarrow w }}{{\left| {\overrightarrow v } \right|^2 }} = \frac{{v_1 w_1 + v_2 w_2 + v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }} = \frac{{v_1 w_1 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_2 w_2 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }} \\ \\ \frac{{\overrightarrow v \cdot \overrightarrow w }}{{\left| {\overrightarrow v } \right|^2 }}\overrightarrow v = \left( {\frac{{v_1 w_1 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_2 w_2 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }}} \right)\left\langle {v_1 ,\,v_2 ,\,v_3 } \right\rangle \\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{v_1 w_1 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_2 w_2 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }}} \right)\left\langle {v_1 ,\,v_2 ,\,v_3 } \right\rangle \\ \end{array}$$

Things are starting to get real ugly. Am I missing an easier way?

2. Feb 13, 2008

Dick

Yes, you are. You want to show v.(w-(v.w)*v/(v.v))=0. Just multiply the outer dot product through.