Question

If a unit vector $\overrightarrow{a}$ makes an angle $\frac{\pi }{3}$ 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 $\overrightarrow{a}$.

Answer 1

Let unit vector $\overrightarrow{a}$ have $(a_1,a_2,a_3)$ components.

$\therefore \overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$

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

Also, it is given that $\overrightarrow{a}$ 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:

$\cos \frac{\pi }{3}=\frac{a_1}{|\overrightarrow{a}|}$

$\Rightarrow \frac{1}{2}=a_1[|\overrightarrow{a}|=1]$

$\cos \frac{\pi }{4}=\frac{a_1}{|\overrightarrow{a}|}$

$\Rightarrow \frac{1}{\sqrt{2}}=a_2[|\overrightarrow{a}|=1]$

Also, $\cos \theta =\frac{a_3}{|\overrightarrow{a}|}$.

$\Rightarrow a_3=\cos \theta$

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 $\overrightarrow{a}$ are $(\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2})$.