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

Let $\quad \overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-\beta \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \quad$ 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 at $\overline{\mathrm{a}}$ on $\overline{\mathrm{c}}$ is $\frac{10}{3}$ and $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$, then the value of $\alpha^2+\beta^2-\alpha \beta$ is equal to
  1. 1
  2. 2
  3. 3
  4. 4

Solution

Projection of $\overline{\mathrm{a}}$ on $\overline{\mathrm{c}}$ is $\frac{10}{3}$ $\begin{aligned} & \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{|\overline{\mathrm{c}}|}=\frac{10}{3} \\ & \Rightarrow \frac{(\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{1+4+4}}=\frac{10}{3} \\ & \Rightarrow \frac{\alpha+3(2)-1(-2)}{\sqrt{9}}=\frac{10}{3} \\ & \Rightarrow \frac{\alpha+8}{3}=\frac{10}{3} \\ & \Rightarrow \alpha=2 \\ & \overline{\mathrm{~b}} \times \overline{\mathrm{c}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & \Rightarrow\left|\begin{array}{cc} \hat{\mathrm{i}} & \hat{\mathrm{j}} \quad \hat{\mathrm{k}} \\ 3 & -\beta \\ 1 & \quad 2 \end{array} \quad-2\right|=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & \Rightarrow(2 \beta-8) \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+(6+\beta) \hat{\mathrm{k}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & \Rightarrow 6+\beta=7 \Rightarrow \beta=1 \\ & \alpha^2+\beta^2-\alpha \beta=4+1-2=3 \end{aligned}$

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

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