JEE Main 2025 Jan 23 Shift 2, Mathematics Q7: Scalar (or Dot) Product of Two Vectors

Section formula: $A=\left(\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}\right)$.

Compute $\overrightarrow{OQ}\cdot \overrightarrow{OA}$:

$\overrightarrow{OQ}\cdot \overrightarrow{OA}=5\cdot \frac{5r-1}{r+1}+5\cdot \frac{5r-1}{r+1}+10\cdot \frac{10r+2}{r+1}=\frac{5(30r+2)}{r+1}\cdot ...$

Simplified: $\overrightarrow{OQ}\cdot \overrightarrow{OA}=\frac{5(30r+2)}{r+1}\cdot \frac{1}{5}\cdot 5=\frac{5(30r+2)}{r+1}$. (Equivalently, $\frac{10(15r+1)}{r+1}$.)

For the cross product term, observe that $\overrightarrow{OA}=\frac{r\overrightarrow{OQ}+\overrightarrow{OP}}{r+1}$, so $\overrightarrow{OP}\times \overrightarrow{OA}=\frac{r}{r+1}\overrightarrow{OP}\times \overrightarrow{OQ}$. Compute $\overrightarrow{OP}\times \overrightarrow{OQ}$ with $\overrightarrow{OP}=(-1,-1,2)$, $\overrightarrow{OQ}=(5,5,10)$:

$\overrightarrow{OP}\times \overrightarrow{OQ}=(-10-10,10-(-10),-5-(-5))=(-20,20,0).$

$|\overrightarrow{OP}\times \overrightarrow{OQ}{|}^2=400+400+0=800$, so $|\overrightarrow{OP}\times \overrightarrow{OA}{|}^2=\frac{800r^2}{(r+1{)}^2}$.

Plug into the given equation:

$\frac{10(15r+1)}{r+1}-\frac{1}{5}\cdot \frac{800r^2}{(r+1{)}^2}=10.$

Multiply through by $(r+1{)}^2$ and simplify to $2r^2-14r=0\Rightarrow r(r-7)=0$. Since $r>0$, $r=7$.