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

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(A) Let the unit vector $\vec{a}$ be $\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$.Since $\vec{a}$ is a unit vector,$|\vec{a}| = 1$.The direc

Vedclass · Accepted Answer

$	heta = \frac{\pi}{3}, \vec{a} = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}ight)$

Vedclass · Answer

(A) Let the unit vector $\vec{a}$ be $\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$.Since $\vec{a}$ is a unit vector,$|\vec{a}| = 1$.The direction cosines of $\vec{a}$ are $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha = \frac{\pi}{3}, \beta = \frac{\pi}{4}$,and $\gamma = 	heta$.Thus,$a_{1} = \cos \frac{\pi}{3} = \frac{1}{2}$,$a_{2} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $a_{3} = \cos 	heta$.We know that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.Substituting the values: $\left(\frac{1}{2}ight)^{2} + \left(\frac{1}{\sqrt{2}}ight)^{2} + \cos^{2} 	heta = 1$.$\frac{1}{4} + \frac{1}{2} + \cos^{2} 	heta = 1$.$\frac{3}{4} + \cos^{2} 	heta = 1 \Rightarrow \cos^{2} 	heta = \frac{1}{4}$.Since $	heta$ is acute,$\cos 	heta = \frac{1}{2}$,which implies $	heta = \frac{\pi}{3}$.Therefore,the components of $\vec{a}$ are $\left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}ight)$.

If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$,$\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$,then find $\theta$ and the components of $\vec{a}$.

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