Question

if $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$

Answer 1

$\overline{a}.\overline{b}=\overline{b}.\overline{c}=\overline{c}.\overline{a}=0$

Let $\overline{a}+\overline{b}+\overline{c}=\overline{d}$ and $\overline{d}$ makes angle $\alpha ,\beta ,\gamma$ with $\overline{a},\overline{b},\overline{c}$

$cos\alpha =\frac{\overline{v},\overline{a}}{|v||\overline{a}|},cos\beta =\frac{\overline{v}.\overline{b}}{|v||\overline{b}|},cos\gamma =\frac{\overline{v}.\overline{c}}{|v||\overline{c}|}$

$cos\alpha =\frac{(\overline{a}+\overline{b}+\overline{c}).\overline{a}}{|v||\overline{a}|}=\frac{\overline{a}.\overline{a}+\overline{a}.\overline{b}+\overline{a}.\overline{c}}{|v||\overline{a}|}=\frac{|a{|}^2}{|v||\overline{a}|}=\frac{|a|}{|v|}$

$cos\beta =\frac{|b|}{|v|}andcos\gamma =\frac{|c|}{|v|}$

given that $|\overline{a}|=|\overline{b}|=|\overline{c}|$

hence $cos\alpha =cos\beta =cos\gamma$

$\Rightarrow \alpha =\beta =\gamma$

$\therefore \overline{a}+\overline{b}+\overline{c}$ is equal inclined to $\overline{a},\overline{b}$ and $\overline{c}$