QJEEM25Ma5a90: JEE Main 2025 Physics - Class 11 JEEM25Ma5a90 Solution: Let and be a vector such that and . Then the maximum value of…

$\begin{aligned}& \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{~b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\& \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} \\& \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=0 \\& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\& \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\& \overrightarrow{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \ldots . .(1) \\& |\overrightarrow{\mathrm{c}}|^2=\lambda^2(25+36+16) \\& |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168 \\& \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~b}}+|\overrightarrow{\mathrm{c}}|^2=168 \\& 14+\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})+77 \lambda^2=168\end{aligned}$

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 $\lambda=-2$

$\begin{aligned}& |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\& =77 \times 4 \\& =308\end{aligned}$