Let $\bar{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k} \quad$ and…

Let $\bar{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k} \quad$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$, where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection of $\overline{\mathrm{a}}$ on $\overline{\mathrm{c}}$ is $\frac{10}{3}$ and $\bar{b} \times \bar{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$, then the value of $2 \alpha+\beta$ is
  1. 3
  2. 4
  3. 5
  4. 6

Solution

Projection of $\overline{\mathrm{a}}$ on $\overline{\mathrm{c}}=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{|\overline{\mathrm{c}}|}$ $\begin{aligned} & \therefore \quad \frac{\alpha+6+2}{\sqrt{1+4+4}}=\frac{10}{3} \\ & \therefore \quad \alpha=2 \\ & \overline{\mathbf{b}} \times \overline{\mathbf{c}}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{array}\right| \\ & =(2 \beta-8) \hat{i}-(-6-4) \hat{j}+(6+\beta) \hat{k} \\ & =(2 \beta-8) \hat{i}+10 \hat{j}+(6+\beta) \hat{k} \end{aligned}$
Comparing with $-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$, we get $\begin{array}{ll} & 2 \beta-8=-6 \\ \therefore & 2 \beta=2 \\ \therefore \quad & \beta=1 \\ \therefore \quad & 2 \alpha+\beta=5 \end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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