Question

If $\hat{i},\hat{j}$ and $\hat{k}$ are unit vectors along x, y and z – axes respectively, the angle $\theta$ between the vector $\hat{i}+\hat{j}+\hat{k}$ and vector $\hat{i}$ is given by

• A. $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  
• B. $\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  
• C. $\theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
  
• D. $\theta ={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
  

Answer 1

The correct option is A $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$


$\cos \theta =\frac{(\hat{i}+\hat{j}+\hat{k}).\hat{i}}{(\hat{i}{}^2+\hat{j}{}^2+\hat{k}{}^2{)}^{1/2}\times \hat{i}}=\frac{\hat{i}.\hat{i}+\hat{j}.\hat{i}+\hat{k}.\hat{i}}{(1+1+1{)}^{1/2}\times 1}=\frac{1+0+0}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$(\because \hat{j}.\hat{i}+\hat{k}.\hat{i}=0$ and $\hat{i}.\hat{i}=1)$
Hence the correct choice is (a).