Question

If $\overline{x},\overline{y},\overline{z}$ are mutually perpendicular vectors of eqaul magnitude, then find the measure of an angle that $\overline{x}+\overline{y}+\overline{z}$ makes with any of the three vectors.

Answer 1

Assuming,

$\begin{array}{c}\overline{x}\perp \overline{y}\perp \overline{z}andequalmagnitudeis\left|\overline{x}\right|=\left|\overline{y}\right|=\left|\overline{z}\right|=k\\ Lets,\\ \left(\overline{x}+\overline{y}+\overline{z}\right).\overline{x}=\left|\overline{x}+\overline{y}+\overline{z}\right|\cdot \left|\overline{x}\right|.\cos \theta \\ simplifyof,\\ \left|\overline{x}\right|.\overline{x}+\overline{x}.\overline{y}+\overline{x}.\overline{z}=\left|\overline{x}+\overline{y}+\overline{z}\right|\cdot \left|\overline{x}\right|.\cos \theta \\ {\left|\overline{x}\right|}^2+0+0=\left|\overline{x}+\overline{y}+\overline{z}\right|\cdot \left|\overline{x}\right|.\cos \theta \\ weknowthat,\\ k^2=\left|\overline{x}+\overline{y}+\overline{z}\right|\cdot k.\cos \theta \\ \cos \theta =\frac{k}{\left|\overline{x}+\overline{y}+\overline{z}\right|}\\ letssquringof\left|\overline{x}+\overline{y}+\overline{z}\right|\end{array}$

$\begin{array}{c}{\left|\overline{x}+\overline{y}+\overline{z}\right|}^2={\left|\overline{x}\right|}^2+{\left|\overline{y}\right|}^2+{\left|\overline{z}\right|}^2+2\left|\overline{x}\right|\left|\overline{y}\right|+2\left|\overline{y}\right|\left|\overline{z}\right|+2\left|\overline{x}\right|\left|\overline{z}\right|\\ =k^2+k^2+k^2+0+0+0\\ {\left|\overline{x}+\overline{y}+\overline{z}\right|}^2=3k^2\\ \therefore \left|\overline{x}+\overline{y}+\overline{z}\right|=\sqrt{3}k\\ wehave,\\ \cos \theta =\frac{k}{\left|\overline{x}+\overline{y}+\overline{z}\right|}\\ \cos \theta =\frac{k}{\sqrt{3}k}=\frac{1}{\sqrt{3}}\\ \therefore \theta ={\cos }^{-1}\frac{1}{\sqrt{3}}\end{array}$