Question
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 hence, the components of $\vec a $.

Answer

Let $\vec{\mathrm{a}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$ 
Given that it is a unit vector,
So, $|\vec{\mathrm{a}}|=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}}=1$ ......(i)
Let $\alpha, \beta, \theta$ be the angles  $\vec a$ makes with $\hat{i}, \hat{\jmath}, \hat{\mathrm{k}}$ respectively.
Then,
$\cos \alpha=\frac{\vec{a} \cdot \hat{\imath}}{|\vec{a}||\hat{\imath}|}=\frac{x}{1}$ 
$\Rightarrow \alpha=\frac{\pi}{3}$ (Given)
$\Rightarrow x=\cos \alpha=\cos \frac{\pi}{3}=\frac{1}{2}$ 
$\cos \beta=\frac{\vec{\mathrm{a}} \cdot \hat{\mathrm{j}}}{|\vec{\mathrm{a}}||\hat{\mathrm{j}}|}=\frac{\mathrm{y}}{1}$ 
$\Rightarrow \beta=\frac{\pi}{4}$ (Given)
$\Rightarrow \mathrm{y}=\cos \beta=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ 
$\cos \theta=\frac{\vec{\mathrm{a}} \cdot \hat{\mathrm{k}}}{|\vec{\mathrm{a}}||\hat{\mathrm{k}}|}=\frac{\mathrm{z}}{1}$ 
$\Rightarrow {z}=\cos \theta$ 
Putting value of x,y, z in equation (i)
$\sqrt{x^{2}+y^{2}+z^{2}}=1$ 
$\Rightarrow x^{2}+y^{2}+z^{2}=1$ (Squaring both sides)
$\Rightarrow\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+(\cos \theta)^{2}=1$ 
$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \theta=1$ 
$\Rightarrow \cos ^{2} \theta=1-\frac{3}{4}=\frac{1}{4}$ 
$\Rightarrow \cos \theta=\pm \frac{1}{2}$ 
As $\theta$ should be acute, $\theta=\cos ^{-1} \frac{1}{2}=\frac{\pi}{3}$ 
$\therefore \mathrm{z}=\cos \theta=\cos \frac{\pi}{3}=\frac{1}{2}$ 
Components of $\vec{\mathrm{a}}$ are the coefficients of $\hat{\imath}, \hat{\jmath}, \hat{\mathrm{k}}$ which are 
$x=\frac{1}{2} ; y=\frac{1}{\sqrt{2}} ; z=\frac{1}{2}$ with $\theta=\frac{\pi}{3}$ 

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