# If a unit vector $$\overrightarrow a$$ makes angles $$\frac{\large \pi}{\large 3}$$ with $$\hat i, \frac{\large \pi}{\large 3}$$ with $$\hat j$$ and an acute angle $$θ$$ with $$\hat k$$ , then find $$θ$$ and hence, the components of $$\overrightarrow a$$ .

Toolbox:
• Sum of the squares of the direction cosines is $\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3=1$
Step 1:
It is given that $\overrightarrow a$ makes an angle $\large\frac{\pi}{3}$ with $\hat i$ and $\large\frac{\pi}{4}$ with $\hat j$ and acute angle with $\hat k$
Let us find the acute angle $\theta$ made with $\hat k$
We know sum of the squares of the direction cosines of angles made with $\hat i,\hat j$ and $\hat k$ is 1.
(i.e) $\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3$=1.
Here $\cos\theta_1=\cos\large\frac{\pi}{3}=\large\frac{1}{2}$
$\quad\cos\theta_2=\cos\large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}$
Let the third angle be $\theta_3$
Therefore $\big(\large\frac{1}{2}\big)^2+\big(\large\frac{1}{\sqrt 2}\big)^2+$$\cos^2\theta_3=1 Step 2: \cos^2\theta_3=1-\big(\large\frac{1}{4}+\large\frac{1}{2}\big) \qquad\quad=1-\big(\large\frac{3}{4}\big) \qquad\quad=\big(\large\frac{1}{4}\big) \cos^2\theta_3=\large\frac{1}{4}\Rightarrow$$\cos\theta_3=\pm\large\frac{1}{2}$
Step 3:
But it is said that $\theta_3$ is an acute angle.
Therefore $\cos\theta_3=\large\frac{1}{2}$
$\Rightarrow \theta_3=\cos^{-1}\big(\large\frac{1}{2}\big)$
$\qquad\;=\large\frac{\pi}{3}$
Step 4:
Hence the angle made with $\hat k$ is $\large\frac{\pi}{3}$
The components of $\overrightarrow a$ are $\large\frac{1}{2},\frac{1}{\sqrt 2}$ and $\large\frac{1}{2}$