If $\overline{\mathrm{a}} \cdot$ and $\overline{\mathrm{c}}$ are unit vectors inclined at $\frac{\pi}{3}$…
- -10
- 10
- 50
- -50
Solution
Now, $[\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})] \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=5$ ...[Given] $\Rightarrow[(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \overline{\mathrm{c}}](\overline{\mathrm{a}} \times \overline{\mathrm{c}})=5$ $\begin{aligned} & \Rightarrow\left[\frac{1}{2} \overline{\mathrm{~b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}\right] \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=5 \\ & \Rightarrow \frac{1}{2} \overline{\mathrm{~b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=5 \\ & \Rightarrow \overline{\mathrm{~b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=10 \\ & \Rightarrow[\overline{\mathrm{~b}} \overline{\mathrm{a}} \overline{\mathrm{c}}]=10 \\ & \Rightarrow-[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=10 \\ & \Rightarrow[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=-10 \\ & \Rightarrow 5[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=-50\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 1)