Question

The angle between the vectors $(2,-1,1)$ and $(1,-1,2)$ is ________

• A. ${\sin }^{-1}\frac{\sqrt{11}}{6}$
• B. ${\sin }^{-1}\frac{5}{6}$
• C. $\frac{\pi }{2}$
• D. ${\cos }^{-1}\frac{1}{6}$

Answer 1

Answer: (A)

${\sin }^{-1}\frac{\sqrt{11}}{6}$

First vector is $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}$

$⟹a=\sqrt{2^2+1^2+1^2}=\sqrt{6}$

Second vector is $\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}$

$⟹b=\sqrt{1^2+1^2+2^2}=\sqrt{6}$

Now, we know that

$\overrightarrow{a}.\overrightarrow{b}=ab\cos \theta$

$⟹(2\hat{i}-\hat{j}+\hat{k})(\hat{i}-\hat{j}+2\hat{k})=6\cos \theta$

$⟹\cos \theta =\frac{5}{6}$

$⟹\sin \theta =\sqrt{1-{\cos }^2\theta }=\frac{\sqrt{11}}{6}$

$⟹\theta ={\sin }^{-1}\frac{\sqrt{11}}{6}$