If $\overline{\mathrm{a}} \cdot$ and $\overline{\mathrm{c}}$ are unit vectors inclined at $\frac{\pi}{3}$…

If $\overline{\mathrm{a}} \cdot$ and $\overline{\mathrm{c}}$ are unit vectors inclined at $\frac{\pi}{3}$ with each other and $(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})) \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})=5$, then the value of $5[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=$
  1. -10
  2. 10
  3. 50
  4. -50

Solution

$\begin{aligned} |\overline{\mathrm{a}}|=|\overline{\mathrm{c}}| & =1 \\ \overline{\mathrm{a}} \cdot \overline{\mathrm{c}} & =|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \cos \frac{\pi}{3} \\ & =1 \times 1 \times \frac{1}{2} \\ & =\frac{1}{2} \end{aligned}$
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)

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