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)$