Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, then the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is given by
Options:
Solution:
2929 Upvotes
Verified Answer
The correct answer is:
$\sec ^{-1}(1)$
$\begin{array}{l}
\cos \theta=\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \\
=\frac{1 \times 2+(-1) \times(-1)+2 \times(1)}{\sqrt{1+1+4} \sqrt{4+1+1}} \\
=\frac{2+2+2}{6}=\frac{6}{6}=1
\end{array}$
So, $\theta=0^{\circ}$ or $\theta=2 \pi$
$\begin{array}{ll}\because & \sec 2 \pi=1 \\ \therefore & 2 \pi=\sec ^{-1}(1) \\ \Rightarrow & \theta=\sec ^{-1}(1)\end{array}$
\cos \theta=\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \\
=\frac{1 \times 2+(-1) \times(-1)+2 \times(1)}{\sqrt{1+1+4} \sqrt{4+1+1}} \\
=\frac{2+2+2}{6}=\frac{6}{6}=1
\end{array}$
So, $\theta=0^{\circ}$ or $\theta=2 \pi$
$\begin{array}{ll}\because & \sec 2 \pi=1 \\ \therefore & 2 \pi=\sec ^{-1}(1) \\ \Rightarrow & \theta=\sec ^{-1}(1)\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.