If a unit vector makes an angles

Let unit vector $\vec{a}$ have $\left(a_{1}, a_{2}, a_{3}\right)$ components.

$\Rightarrow \vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$

Since $\vec{a}$ is a unit vector, $|\vec{a}|=1$.

Also, it is given that $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}, \frac{\pi}{4}$ with $\hat{j}$, and an acute angle $\theta$ with $\hat{k}$.

Then, we have:

$\cos \frac{\pi}{3}=\frac{a_{1}}{|\vec{a}|}$

$\Rightarrow \frac{1}{2}=a_{1} \quad[|\vec{a}|=1]$

$\cos \frac{\pi}{4}=\frac{a_{2}}{|\vec{a}|}$

$\Rightarrow \frac{1}{\sqrt{2}}=a_{2} \quad[|\vec{a}|=1]$

Also, $\cos \theta=\frac{a_{3}}{|\vec{a}|}$

$\Rightarrow a_{3}=\cos \theta$

Now,

$|a|=1$

$\Rightarrow \sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}=1$

$\Rightarrow\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\cos ^{2} \theta=1$

$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \theta=1$

$\Rightarrow \frac{3}{4}+\cos ^{2} \theta=1$

$\Rightarrow \cos ^{2} \theta=1-\frac{3}{4}=\frac{1}{4}$

$\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$

$\therefore a_{3}=\cos \frac{\pi}{3}=\frac{1}{2}$

Hence, $\theta=\frac{\pi}{3}$ and the components of $\vec{a}$ are $\left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)$.