Q33 / 75 · JEE Main 2025 (29 Jan Shift 1) · Mathematics

Let \(\overrightarrow{\mathrm{a}}=2 \hat{i}-\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{~b}}=3 \hat{i}-5 \hat{j}+\hat{k}\) and \(\overrightarrow{\mathrm{c}}\) be a vector such that \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}\) and \((\vec{a}+\vec{c}) \cdot(\vec{b}+\vec{c})=168\). Then the maximum value of \(|\vec{c}|^2\) is :

  • A
    462
  • B
    77
  • C
    154
  • Correct answer
    D
    308

Solution

$\begin{aligned} & \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ & \overrightarrow{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \ldots . .(1) \\ & |\overrightarrow{\mathrm{c}}|^2=\lambda^2(25+36+16) \\ & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168 \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~b}}+|\overrightarrow{\mathrm{c}}|^2=168 \end{aligned}$ $14+\vec{c} \cdot(\vec{a}+\vec{b})+77 \lambda^2=168$ using equation (1) $\begin{aligned} & \lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^2+77 \lambda^2=154 \\ & 77 \lambda+77 \lambda^2-154=0 \\ & \lambda^2+\lambda-2=0 \\ & \lambda=-2,1 \end{aligned}$ $\therefore$ Maximum value of $|\overrightarrow{\mathrm{c}}|^2$ occurs when $\lambda=-2$ $\begin{aligned} & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & =77 \times 4 \\ & =308 \end{aligned}$

Hence, the correct answer is option: 'D'.

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