Question

What is the angle between the following vectors?

$\overrightarrow{A}=3\hat{i}-2\hat{j}+\hat{k}$

$\overrightarrow{B}=2\hat{i}+6\hat{j}-6\hat{k}$

• A. $co{s}^{-1}\left(\frac{12}{\sqrt{266}}\right)$
• B. $co{s}^{-1}\left(\frac{-6}{\sqrt{266}}\right)$
• C. $co{s}^{-1}\left(\frac{-6}{\sqrt{1064}}\right)$
• D. None of these

Answer 1

The correct option is B

$co{s}^{-1}\left(\frac{-6}{\sqrt{266}}\right)$

$\overrightarrow{A}.\overrightarrow{B}=(3\times 2)+(-2\times 6)+(1\times -6)$

= 6 - 12 - 6

= -12

$|\overrightarrow{A}|=\sqrt{3^2+(-2{)}^2+1^2}=\sqrt{9+4+1}=\sqrt{14}$

$|\overrightarrow{B}|=\sqrt{2^2+6^2+(-6{)}^2}=\sqrt{4+36+36}=\sqrt{76}$

$\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}=\frac{-12}{\sqrt{14}\sqrt{76}}=\frac{-6}{\sqrt{266}}$

$\therefore cos\theta =\frac{-6}{\sqrt{266}}$

$\Rightarrow \theta =co{s}^{-1}\left(\frac{-6}{\sqrt{266}}\right)$