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

Answer

Let unit vector $\vec{\text{a}}$ have $(a_1, a_2, a_3)$ components.
$\Rightarrow\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
Since $\vec{\text{a}}$ is a unit vector, $|\vec{\text{a}}|=1$
Also it is given that $\vec{\text{a}}$ makes angles $\frac{\pi}3$ with $\hat{\text{i}}$, $\frac{\pi}4$ with $\hat{\text{j}}$, and an acute angle $\theta$ with $\hat{\text{k}}$.
Then, we have:
$\cos\frac{\pi}3=\frac{\text{a}_1}{|\vec{\text{a}}|}$
$\Rightarrow\ \frac{1}2=\text{a}_1$ $[|\vec{\text{a}}|=1]$
$\cos\frac{\pi}4=\frac{\text{a}_2}{|\vec{\text{a}}|}$
$\Rightarrow\ \frac{1}{\sqrt2}=\text{a}_2$ $[|\vec{\text{a}}|=1]$
Also, $\cos\theta=\frac{\text{a}_3}{|\vec{\text{a}}|}$
$\Rightarrow\ \text{a}_3=\cos\theta$
Now,
$|\text{a}|=1$
$\Rightarrow\ \sqrt{\text{a}_1^2+\text{a}_2^2+\text{a}_3^2}=1$
$\Rightarrow\Big(\frac{1}2\Big)^2+\Big(\frac{1}{\sqrt{2}}\Big)^2+\cos^2\theta=1$
$\Rightarrow\ \frac{1}4+\frac{1}2+\cos^2\theta=1$
$\Rightarrow\ \frac{3}4+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\frac{3}4=\frac{1}4$
$\Rightarrow\ \cos\theta=\frac{1}2$
$\Rightarrow\ \theta=\frac{\pi}3$
$\therefore\ \text{a}_3=\cos\frac{\pi}3=\frac{1}2$
Hence, $\theta=\frac{\pi}3$ and the components of $\vec{\text{a}}$ are $\Big(\frac{1}2,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\Big)$.

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