Question

If $\hat{i},\hat{j}$ and $\hat{k}$ represent unit vectors along the $x,y$ and $z$-axes respectively, then the angle $\theta$ between the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}$ is equal to

• A. ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. ${\sin }^{-1}\left(\sqrt{\frac{2}{3}}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. $90^{\circ }$

Answer 1

Answer: (A)

Let $\vec{A}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{B}=\hat{i}+\hat{j}$
and angle between $\vec{A}$ and $\vec{B}$ be $\theta$
As $\vec{A}\cdot \vec{B}=AB\cos \theta$
$\Rightarrow \cos \theta =\frac{\vec{A}\cdot \vec{B}}{AB}$... (i)
Now, $|\vec{A}|=\sqrt{(1{)}^2+(1{)}^2+(1{)}^2}=\sqrt{3}$
and $\vec{B}=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}$
$\therefore \cos \theta =\frac{(\hat{i}+\hat{j}+\hat{k})\cdot (\hat{i}+\hat{j})}{\sqrt{3}\cdot \sqrt{2}}=\frac{1+1}{\sqrt{6}}=\sqrt{\frac{2}{3}}$
and $\sin \theta =\sqrt{1-{\cos }^2\theta }$
or $\sin \theta =\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}}$