Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors of equal magnitude, show that the vectors $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is equally inclined to $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$.

Answer 1

To prove:$\overline{a}+\overline{b}+\overline{c}$ is equally inclined to $\overline{a},\overline{b}and\overline{c}.$

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

Angle between:$\overline{a}+\overline{b}+\overline{c}and\overline{a}:$

$\cos {\theta }_1=\frac{(\overline{a}+\overline{b}+\overline{c}).\overline{a}}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{a}\right|}⟹\cos {\theta }_1=\frac{{\left|\overline{a}\right|}^2+\overline{a}.\overline{b}+\overline{a}.\overline{c}}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{a}\right|}⟹\cos {\theta }_1=\frac{{\left|\overline{a}\right|}^2+0+0}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{a}\right|}=\frac{\left|\overline{a}\right|}{|\overline{a}+\overline{b}+\overline{c}|}$

Angle between $\overline{a}+\overline{b}+\overline{c}and\overline{b}$

$⟹\cos {\theta }_2=\frac{(\overline{a}+\overline{b}+\overline{c}).\overline{b}}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{b}\right|}=\frac{\overline{a}.\overline{b}+{\left|\overline{b}\right|}^2+\overline{b}.\overline{c}}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{b}\right|}=\frac{0+{\left|\overline{b}\right|}^2}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{b}\right|}=\frac{\left|\overline{b}\right|}{|\overline{a}+\overline{b}+\overline{c}|}$

Angle between $\overline{a}+\overline{b}+\overline{c}and\overline{c}:$

$⟹\cos {\theta }_3=\frac{(\overline{a}+\overline{b}+\overline{c}).\overline{c}}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{c}\right|}=\frac{\overline{a}.\overline{c}+\overline{b}.\overline{c}+{\left|\overline{c}\right|}^2}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{c}\right|}=\frac{0+0+{\left|\overline{c}\right|}^2}{|\overline{a}+\overline{b}+\overline{c}|.\left|\overline{c}\right|}=\frac{\left|\overline{c}\right|}{|\overline{a}+\overline{b}+\overline{c}|}\because \left|\overline{a}\right|=\left|\overline{b}\right|=\left|\overline{c}\right|=p\left(let\right)\therefore \cos {\theta }_1=\cos {\theta }_2=\cos {\theta }_3=\frac{p}{|\overline{a}+\overline{b}+\overline{c}|}$

Hence proved.