Question

If the vectors $\overline{a}=2i+j-k$ and $\overline{b}=i+j+k$, then find the angle between its.

• A. ${\sin }^{-1}\frac{\sqrt{2}}{3}$
• B. ${\cos }^{-1}\frac{\sqrt{2}}{3}$
  
• C. ${\cos }^{-1}\frac{2}{\sqrt{3}}$
  
• D. ${\sin }^{-1}\frac{2}{\sqrt{3}}$
  

Answer 1

The correct option is B ${\cos }^{-1}\frac{\sqrt{2}}{3}$

Consider the given vectors.

$a=2i+j-k$

$b=i+j+k$

We know that

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

Therefore,

$\cos \theta =\frac{(2i+j-k).(i+j+k)}{|2i+j-k||i+j+k|}$

$\cos \theta =\frac{(2+1-1)}{\left(\sqrt{4+1+1}\right)\left(\sqrt{1+1+1}\right)}$

$\cos \theta =\frac{2}{\left(\sqrt{6}\right)\left(\sqrt{3}\right)}$

$\cos \theta =\frac{2}{3\sqrt{2}}$

$\cos \theta =\frac{\sqrt{2}}{3}$

$\theta ={\cos }^{-1}\frac{\sqrt{2}}{3}$

Hence, this is the answer.