Let \(\mathbf{u}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and…
Let \(\mathbf{u}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{v}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\). Then angle between \(\mathbf{u}\) and \(\mathbf{v}\) is
\(\cos ^{-1}\left(\frac{4}{9}\right)\)
\(\cos ^{-1}\left(\frac{-4}{3}\right)\)
\(\cos ^{-1}\left(\frac{4}{3}\right)\)
\(\sin ^{-1}\left(\frac{-4}{9}\right)\)
Solution
Let angle between vector \(\mathbf{u}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{v}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) is \(\theta\), then
\(\begin{aligned}
\cos \theta & =\frac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}||\mathbf{v}|}=\frac{|-2-4+2|}{\sqrt{4+4+1} \sqrt{1+4+4}}=\frac{4}{9} \\
\Rightarrow \quad \theta & =\cos ^{-1}\left(\frac{4}{9}\right)
\end{aligned}\)
Hence, option (a) is correct.