Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3…

Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\overrightarrow{\mathrm{r}}$ be anit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$, then $3 \lambda$ is equal to:
  1. 21
  2. 30
  3. 25
  4. 27

Solution

$\begin{aligned} & \overrightarrow{\mathrm{r}}=\mathrm{k}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=3 \\ & \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \\ & 3=\mathrm{k}(2+6-15+3-2+3 \lambda)\end{aligned}$ $3=\mathrm{k}(-6+3 \lambda)$ ...(1) $\overrightarrow{\mathrm{r}}=\mathrm{k}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-(5-\lambda) \hat{\mathrm{k}})$ $|\vec{r}|=k \sqrt{25+4+25+\lambda^2-10 \lambda}=1$ ...(2) $\mathrm{k}=\frac{3}{-6+3 \lambda}=\frac{1}{-2+\lambda} \quad$ put in (2) $\begin{aligned} & 4+\lambda^2-4 \lambda=54+\lambda^2-10 \lambda \\ & 6 \lambda=50 \\ & 3 \lambda=25\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 2)

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