JEE Main 2024 — 8 April, Shift 2 — Question 13

$\quad(\vec{a}+\vec{b}) \times \vec{c}-\vec{c} \times(-2 \vec{a}+3 \vec{b})=0$

$(\vec{a}+\vec{b}) \times \vec{c}+(-2 \vec{a}+3 \vec{b}) \times \vec{c}=0$ $\Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{~b}}) \times \overrightarrow{\mathrm{c}}=0$

$\Rightarrow \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{~b}}-\overrightarrow{\mathrm{a}})$

$\Rightarrow\lambda(44 \hat{i}-4 \hat{j}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$ $=\lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})$

Now $(8 \hat{i}-2 \hat{j}+2 \hat{k}+33 \hat{i}-3 \hat{j}+3 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})=1670$

$\Rightarrow(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot(40 \hat{i}-3 \hat{j}+3 \hat{k}) \times \lambda=1670)$

$\Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1$

so $\overrightarrow{\mathrm{c}}=40 \hat{i}-3 \hat{j}-3 \hat{k}$

$\Rightarrow|\vec{c}|^{2}=1600+9+9=1618$