Search any question & find its solution
Question:
Answered & Verified by Expert
If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar unit vectors such that $\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{\sqrt{2}}$, then the angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is
Options:
Solution:
1347 Upvotes
Verified Answer
The correct answer is:
$\frac{3 \pi}{4}$
$\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}}$
$\begin{aligned} & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}} \\ & \Rightarrow\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}-\frac{1}{\sqrt{2}}\right) \overline{\mathrm{b}}-\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\frac{1}{\sqrt{2}}\right) \overline{\mathrm{c}}=0\end{aligned}$
Since $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar unit vectors,
$\begin{aligned} & \bar{a} \cdot \bar{b}+\frac{1}{\sqrt{2}}=0 \\ & \Rightarrow \bar{a} \cdot \bar{b}=-\frac{1}{\sqrt{2}} \\ & \Rightarrow|\bar{a}||\bar{b}| \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \theta=\frac{3 \pi}{4}\end{aligned}$
[Note: In the question, $\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{\sqrt{2}}$ is changed to $\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}}$ to apply appropriate textual concepts.]
$\begin{aligned} & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}} \\ & \Rightarrow\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}-\frac{1}{\sqrt{2}}\right) \overline{\mathrm{b}}-\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\frac{1}{\sqrt{2}}\right) \overline{\mathrm{c}}=0\end{aligned}$
Since $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar unit vectors,
$\begin{aligned} & \bar{a} \cdot \bar{b}+\frac{1}{\sqrt{2}}=0 \\ & \Rightarrow \bar{a} \cdot \bar{b}=-\frac{1}{\sqrt{2}} \\ & \Rightarrow|\bar{a}||\bar{b}| \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \theta=\frac{3 \pi}{4}\end{aligned}$
[Note: In the question, $\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{\sqrt{2}}$ is changed to $\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}}$ to apply appropriate textual concepts.]
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.