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 $\vec{A} \cdot \vec{B}=A B \cos \theta$
$\Rightarrow \cos \theta=\frac{\vec{A} \cdot \vec{B}}{A B}$... (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 $\sin \theta=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}}$