Let $\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and…

Let $\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\bar{a}$ and $\bar{b}$, is $8 \sqrt{3}$ sq. units, then $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}$ is equal to
  1. 1
  2. 2
  3. 3
  4. 4

Solution

$\begin{aligned} \overline{\mathrm{a}} \times \overline{\mathrm{b}} & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -\alpha & 1 \\ 1 & \alpha & 3 \end{array}\right| \\ & =-4 \alpha \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+4 \alpha \hat{\mathrm{k}} \end{aligned}$
Area of parallelogram $=|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|$ $\begin{aligned} & \Rightarrow 8 \sqrt{3}=\sqrt{16 \alpha^2+64+16 \alpha^2} \\ & \Rightarrow 8 \sqrt{3}=\sqrt{32 \alpha^2+64} \end{aligned}$
Squaring on both sides, we get $\begin{aligned} & 192=32 \alpha^2+64 \\ & \Rightarrow 32 \alpha^2=128 \\ & \Rightarrow \alpha^2=4 \\ & \begin{aligned} & \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=(3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \quad=3-\alpha^2+3 \\ & \quad=6-4 \\ & \quad=2 \end{aligned} \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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