Question

If $\overrightarrow{a}=2\hat{i}+4\hat{j}+3\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k},$ then the scalar product and the angle between them are

• A. $13,{\cos }^{-1}\frac{13}{\sqrt{174}}$
• B. $20,{\cos }^{-1}\frac{20}{\sqrt{87}}$
• C. $9,{\cos }^{-1}\frac{9}{\sqrt{87}}$
• D. $9,{\cos }^{-1}\frac{9}{\sqrt{174}}$

Answer 1

The correct option is C $9,{\cos }^{-1}\frac{9}{\sqrt{87}}$

$\overrightarrow{a}.\overrightarrow{b}=a_x{b}_x+a_y{b}_y+a_z{b}_z$
$\overrightarrow{a}=2\hat{i}+4\hat{j}+3\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{a}.\overrightarrow{b}=2+4+3\overrightarrow{a}.\overrightarrow{b}=9$

Angle between them is given as
$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{9}{\sqrt{29}\sqrt{3}}$
$\theta ={\cos }^{-1}\frac{9}{\sqrt{87}}$