Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+5 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-2…

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+5 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be three vectors such that $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$. If $\vec{a} \cdot \vec{c}=-29$, then $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:
  1. 15
  2. 12
  3. 10
  4. 5

Solution

Let's assume $\begin{aligned} \vec{v} & =\vec{a}+\vec{b}+\hat{i} \\ & =5 \hat{i}+3 \hat{j}+\hat{k} \end{aligned}$ and $\overrightarrow{\mathrm{c}}+\hat{\mathrm{i}}=\overrightarrow{\mathrm{p}}$ So, $\begin{aligned} & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{p}} \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0} \\ & \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})=\overrightarrow{0} \\ & \Rightarrow \overrightarrow{\mathrm{p}}=\lambda(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}}) \\ & \overrightarrow{\mathrm{c}}+\mathrm{i}=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \\ & \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{a}} \cdot \hat{\mathrm{i}}=\lambda \overline{\mathrm{a}} \cdot(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \\ & -29+2=\lambda(14+40) \\ & \lambda=-\frac{1}{2} \end{aligned}$ $\begin{gathered}\overrightarrow{\mathrm{c}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\hat{\mathrm{i}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ =-\frac{1}{2}(-14+8)+2=5\end{gathered}$

Asked in: JEE Main 2024 (05 Apr Shift 2)

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