Question

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

Answer 1

The Direction cosines of vector \[\vec{a}\] are \[l = \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}, m = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} , n = \cos \theta\] 
Therefore,  
\[l^2 + m^2 + n^2 = 1\]
\[ \Rightarrow \frac{1}{4} + \frac{1}{2} + n^2 = 1\]
\[ \Rightarrow n^2 = 1 - \frac{3}{4}\]
\[ \Rightarrow n^2 = \frac{1}{4}\]
\[ \Rightarrow n = \frac{1}{2} \left[ \because \vec{a}\text{ makes acute angle with }\hat{k} \right]\]
\[\Rightarrow \cos \theta = \frac{1}{2} \]
\[ \Rightarrow \theta = \frac{\pi}{3}\]
Since, \[\vec{a}\] is the unit vector.
\[\therefore \vec{a} = l \hat{i} + m \hat{j} + n \hat{k} \]
\[ \Rightarrow \vec{a} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}\]
Hence, components of \[\vec{a}\] are \[\frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}\]