Question

The angle between the vectors $(2\hat{i}+6\hat{j}+3\hat{k})$ and $(12\hat{i}-4\hat{j}+3\hat{k})$ is:

• A. ${\cos }^{-1}\left(\frac{1}{10}\right)$
• B. ${\cos }^{-1}\left(\frac{9}{11}\right)$
• C. ${\cos }^{-1}\left(\frac{9}{91}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{9}\right)$

Answer 1

Answer: (C)

Let $\vec{a}=2\hat{i}+6\hat{j}+3\hat{k}$
and $\vec{b}=12\hat{i}-4\hat{j}+3\hat{k}$
$\therefore \cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{(2\hat{i}+6\hat{j}+3\hat{k})\cdot (12\hat{i}-4\hat{j}+3\hat{k})}{\sqrt{4+36+9}\sqrt{144+16+9}}$
$=\frac{24-24+9}{7\cdot 13}=\frac{9}{91}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{9}{91}\right)$