Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such…

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$, then $|\vec{c}|^2$ is equal to :
  1. 1609
  2. 1618
  3. 1600
  4. 1627

Solution

$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}+(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \\ & \Rightarrow=\lambda(44 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =\lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\end{aligned}$ Now $\begin{aligned} & (8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+33 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot \lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=1670 \\ & \Rightarrow(41 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times \lambda=1670) \\ & \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 \\ & \text { so } \overrightarrow{\mathrm{c}}=40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618 \end{aligned}$

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

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