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

Let $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be three vectors. A vector $\overline{\mathrm{v}}$ in the plane of $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, whose projection on $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$, is given by
  1. $\hat{i}-3 \hat{j}+3 \hat{k}$
  2. $-3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
  3. $\quad 3 \hat{i}-\hat{j}+3 \hat{k}$
  4. $\hat{i}+3 \hat{j}-3 \hat{k}$

Solution

$\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+\hat{k} \text { and } \bar{c}=\hat{i}-\hat{j}-\hat{k}$ $\bar{v}$ is in the plane of $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$. $\begin{aligned} & \overline{\mathrm{v}}=\mathrm{m} \overline{\mathrm{a}}+n \overline{\mathrm{~b}} \\ & \Rightarrow \overline{\mathrm{v}}=(m+n) \hat{\mathrm{i}}+(m-n) \hat{j}+(m+n) \hat{\mathrm{k}}...(i) \end{aligned}$
Projection of $\bar{v}$ on $\bar{c}=\frac{\bar{v} \cdot \bar{c}}{|\bar{c}|}=\frac{1}{\sqrt{3}}$ $\begin{aligned} & \Rightarrow \frac{(m+n)(1)+(m-n)(-1)+(m+n)(-1)}{\sqrt{1+1+1}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow-m+n=1 \\ & \Rightarrow \mathrm{n}=1+\mathrm{m} \\ & \mathrm{v}=(2 \mathrm{~m}+1) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(2 \mathrm{~m}+1) \hat{\mathrm{k}} \quad \ldots[\text { From (i) }] \end{aligned}$
When $\mathrm{m}=1$ then $\overline{\mathrm{v}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

Asked in: MHT CET 2024 (15 May Shift 1)

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