If a unit vector vec{r} makes angles frac{pi} | vedclass

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(C) Let the direction angles of the unit vector $\vec{r}$ be $\alpha = \frac{\pi}{3}$,$\beta = \frac{\pi}{4}$,and $\gamma = 	heta$.The sum of the squ

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(C) Let the direction angles of the unit vector $\vec{r}$ be $\alpha = \frac{\pi}{3}$,$\beta = \frac{\pi}{4}$,and $\gamma = 	heta$.The sum of the squares of the direction cosines of a unit vector is always $1$,given by the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Substituting the given values: $\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) + \cos^2 	heta = 1$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 	heta = 1$.$\frac{1}{4} + \frac{1}{2} + \cos^2 	heta = 1$.$\frac{3}{4} + \cos^2 	heta = 1$.$\cos^2 	heta = 1 - \frac{3}{4} = \frac{1}{4}$.Therefore,$\cos 	heta = \pm \frac{1}{2}$.Given $	heta \in (0, \pi)$,if $\cos 	heta = \frac{1}{2}$,then $	heta = \frac{\pi}{3}$. If $\cos 	heta = -\frac{1}{2}$,then $	heta = \frac{2\pi}{3}$.Comparing with the options,$\frac{2\pi}{3}$ is the correct value.

If a unit vector $\vec{r}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$,$\frac{\pi}{4}$ with $\hat{j}$ and $\theta \in (0, \pi)$ with $\hat{k}$,then a value of $\theta$ is

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