JEE Main 2024 — 5 April, Shift 2 — Question 6

Let's assume $\vec{v}=\vec{a}+\vec{b}+\hat{i}$

$$=5 \hat{i}+3 \hat{j}+\hat{k}$$

and $\overrightarrow{\mathrm{c}}+\hat{\mathrm{i}}=\overrightarrow{\mathrm{p}}$

So, $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{p}}$

$\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0}$ $\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})=\overrightarrow{0}$

$\Rightarrow \overrightarrow{\mathrm{p}}=\lambda(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})$ $\overrightarrow{\mathrm{c}}+\mathrm{i}=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}})$ $\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{a}} \cdot \hat{\mathrm{i}}=\lambda \overline{\mathrm{a}} \cdot(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}})$ $-29+2=\lambda(14+40)$

$\lambda=-\frac{1}{2}$

$\overrightarrow{\mathrm{c}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\hat{\mathrm{i}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})$

$=-\frac{1}{2}(-14+8)+2=5$