Question

The angle between the vectors $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}$ is?

• A. ${\tan }^{-1}\frac{1}{2\sqrt{2}}$
• B. ${\tan }^{-1}\frac{1}{3}$
• C. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• D. ${90}^o$

Answer 1

Answer: (A)

${\tan }^{-1}\frac{1}{2\sqrt{2}}$

$\cos \theta =\frac{\overline{a}.\overline{b}}{\left|\overline{a}\right|.\left|\overline{b}\right|}$

$⟹\cos \theta =\frac{(\stackrel{\wedge }{i}+\stackrel{\wedge }{j}+\stackrel{\wedge }{k}).(\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+\stackrel{\wedge }{k})}{\sqrt{3}.\sqrt{1+2^2+1}}$

$⟹\cos \theta =\frac{1+2+1}{\sqrt{3}.\sqrt{6}}$

$⟹\cos \theta =\frac{4}{3\sqrt{2}}$

$⟹\tan \theta =\frac{\sqrt{2}}{2\times 2}=\frac{1}{2\sqrt{2}}$

$⟹\theta ={\tan }^{-1}\left(\frac{1}{2\sqrt{2}}\right)$