Question

Find angle ${}^{\prime }{\theta }^{\prime }$between the vectors $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$.

Answer 1

It is given that $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$
Magnitude of $\overrightarrow{a}=\sqrt{1^2+2^2+(-1{)}^2}=\sqrt{3}$
Magnitude of
$\overrightarrow{b}=\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}$
Now,
$\overrightarrow{a}.\overrightarrow{b}=(\hat{i}+\hat{j}-\hat{k}).(\hat{i}-\hat{j}-\hat{k})$
$\overrightarrow{a}.\overrightarrow{b}=1.1+1.(-1)+(-1)1$
$\overrightarrow{a}.\overrightarrow{b}=-1$
The angle between two vectors is
$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\begin{array}{c}\overrightarrow{a}\end{array}\right|\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|}$
$\cos \theta =\frac{-1}{\sqrt{3}.\sqrt{3}}$
$\cos \theta =\frac{1}{3}$
$\theta ={\cos }^{-1}\left(\frac{-1}{3}\right)$

Final answer:
Hence, the angle between the two vectors is ${\cos }^{-1}\left(\frac{-1}{3}\right)$