- #1

- 1,752

- 143

^{3}. Prove that [tex]\overrightarrow w - {\rm{proj}}_{\overrightarrow v } \overrightarrow w [/tex] is perpendicular to [tex]\overrightarrow v [/tex] .

Here's my attempt:

[tex]\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}[/tex]

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