JEE Main 2026 Jan 28 Shift 1, Mathematics Q25: Applications of Vectors

Since $|\overrightarrow{PR}|=9$: $a^2+b^2+16=81\Rightarrow a^2+b^2=65$. ...(1)

S is equidistant from PQ and PR, so $\overrightarrow{PS}$ is along the angle bisector at P.

$|\overrightarrow{PQ}|=\sqrt{4+1+4}=3$.

Using $\cos {\theta }_1=\cos {\theta }_2$ between $\overrightarrow{PS}$ and $\overrightarrow{PQ}$, $\overrightarrow{PS}$ and $\overrightarrow{PR}$:

$\frac{\overrightarrow{PQ}\cdot \overrightarrow{PS}}{|\overrightarrow{PQ}||\overrightarrow{PS}|}=\frac{\overrightarrow{PR}\cdot \overrightarrow{PS}}{|\overrightarrow{PR}||\overrightarrow{PS}|}$, i.e. $\frac{\overrightarrow{PQ}\cdot \overrightarrow{PS}}{|\overrightarrow{PQ}|}=\frac{\overrightarrow{PR}\cdot \overrightarrow{PS}}{|\overrightarrow{PR}|}$.

$\overrightarrow{PQ}\cdot \overrightarrow{PS}=-2+7+4=9$, so LHS $=9/3=3$.

$\overrightarrow{PR}\cdot \overrightarrow{PS}=a-7b-8$, so RHS $=\frac{a-7b-8}{9}$.

Setting them equal: $a-7b-8=27\Rightarrow a-7b=35$. ...(2)

Solve (1) and (2): from (2), $a=7b+35$. Substitute into (1): $(7b+35{)}^2+b^2=65$, i.e. $49b^2+490b+1225+b^2=65$, $50b^2+490b+1160=0$, $5b^2+49b+116=0$. Discriminant $=49^2-20\cdot 116=2401-2320=81$. Roots: $b=\frac{-49\pm 9}{10}$, giving $b=-4$ or $b=-29/5$ (non-integer).

Take $b=-4$, then $a=7$. $3a-4b=21+16=37$.