JEE Main 2025 — 23 January, Evening Shift — Question 7

Let $A$ divide $P(-1,-1,2)$ and $Q(5,5,10)$ in ratio $r:1$. Coordinates of $A$ are: $\left(\frac{5r-1}{r+1}, \frac{5r-1}{r+1}, \frac{10r+2}{r+1}\right)$. Vectors: $\overrightarrow{OP} = \langle -1, -1, 2 \rangle$, $\overrightarrow{OQ} = \langle 5, 5, 10 \rangle$, $\overrightarrow{OA} = \left\langle \frac{5r-1}{r+1}, \frac{5r-1}{r+1}, \frac{10r+2}{r+1} \right\rangle$. Compute dot product: $\overrightarrow{OQ} \cdot \overrightarrow{OA} = \frac{5(5r-1)+5(5r-1)+10(10r+2)}{r+1} = \frac{150r+10}{r+1} = \frac{10(15r+1)}{r+1}$. Compute cross product magnitude squared: $\overrightarrow{OP} \times \overrightarrow{OA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 2 \\ \frac{5r-1}{r+1} & \frac{5r-1}{r+1} & \frac{10r+2}{r+1} \end{vmatrix}$. $\overrightarrow{OP} \times \overrightarrow{OA}$ can be computed using determinant directly: $\overrightarrow{OP} \times \overrightarrow{OA} = \left( (-1)\cdot\frac{10r+2}{r+1} - 2\cdot\frac{5r-1}{r+1}, 2\cdot\frac{5r-1}{r+1} - (-1)\cdot\frac{10r+2}{r+1}, (-1)\cdot\frac{5r-1}{r+1} - (-1)\cdot\frac{5r-1}{r+1} \right)$. Simplify: First component: $\frac{-10r-2 -10r+2}{r+1} = \frac{-20r}{r+1}$. Second component: $\frac{10r-2+10r+2}{r+1} = \frac{20r}{r+1}$. Third component: 0. Thus, $\overrightarrow{OP} \times \overrightarrow{OA} = \frac{1}{r+1} \langle -20r, 20r, 0 \rangle$. Then $|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = \frac{1}{(r+1)^2} (400r^2+400r^2) = \frac{800r^2}{(r+1)^2}$. Now plug into equation: $\frac{10(15r+1)}{r+1} - \frac{1}{5} \cdot \frac{800r^2}{(r+1)^2} = 10$. Multiply by $5(r+1)^2$: $50(15r+1)(r+1) - 800r^2 = 50(r+1)^2$. Expand: $50(15r^2+16r+1) - 800r^2 = 50(r^2+2r+1)$. $750r^2+800r+50 - 800r^2 = 50r^2+100r+50$. $-50r^2+800r+50 = 50r^2+100r+50$. Bring terms: $-100r^2+700r=0$. $100r^2 - 700r=0$. $100r(r-7)=0$. Since $r>0$, $r=7$.