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

$\quad \overrightarrow{\mathrm{r}}=\mathrm{k}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})$

$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=3$

$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})$

$3=\mathrm{k}(2+6-15+3-2+3 \lambda)$

$3=\mathrm{k}(-6+3 \lambda)$

$\overrightarrow{\mathrm{r}}=\mathrm{k}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-(5-\lambda) \hat{\mathrm{k}})$

$|\overrightarrow{\mathrm{r}}|=\mathrm{k} \sqrt{25+4+25+\lambda^{2}-10 \lambda}=1$

$\mathrm{k}=\frac{3}{-6+3 \lambda}=\frac{1}{-2+\lambda} \quad$ put in (2)

$4+\lambda^{2}-4 \lambda=54+\lambda^{2}-10 \lambda$

$6 \lambda=50$

$3 \lambda=25$