Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ be three non-zero vectors, such that no two of them are collinear and $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the angle between the vectors $\bar{b}$ and $\bar{c}$, then the value of $\sin \theta$ is
Options:
Solution:
1549 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \sqrt{2}}{3}$
Given: $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|^{-}$
We know that,
$(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}^{-}$
On comparing, we get
$\begin{aligned}
& \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\
\Rightarrow & \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \theta \\
\Rightarrow \cos \theta & =\frac{-1}{3} \\
\Rightarrow \cos ^2 \theta & =\frac{1}{9} \\
\sin ^2 \theta & =1-\cos ^2 \theta \\
& =1-\frac{1}{9} \\
\therefore \quad & \sin ^2 \theta=\frac{8}{9} \\
\therefore \quad & \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}$
We know that,
$(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}^{-}$
On comparing, we get
$\begin{aligned}
& \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\
\Rightarrow & \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \theta \\
\Rightarrow \cos \theta & =\frac{-1}{3} \\
\Rightarrow \cos ^2 \theta & =\frac{1}{9} \\
\sin ^2 \theta & =1-\cos ^2 \theta \\
& =1-\frac{1}{9} \\
\therefore \quad & \sin ^2 \theta=\frac{8}{9} \\
\therefore \quad & \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}$
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.