3D Geometry
Internal division of a line segment; dot product condition
nta_pyq_2025_apr
Grade 12
Question:
Let the point $A$ divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $r:1$ ($r>0$). If $O$ is the origin and $\left(\overrightarrow{OQ}\cdot\overrightarrow{OA}\right)-\dfrac{1}{5}\left|\overrightarrow{OP}\times\overrightarrow{OA}\right|^2=10$, then the value of $r$ is:
Step-by-Step Solution
Key Concept: Express $A$ in terms of $r$, compute $\overrightarrow{OQ}\cdot\overrightarrow{OA}$ and $|\overrightarrow{OP}\times\overrightarrow{OA}|^2$ explicitly, substitute into the equation, and solve the resulting quadratic.
$A=\dfrac{(5r-1,5r-1,10r+2)}{r+1}$. $\overrightarrow{OQ}=(5,5,10)$.
$\overrightarrow{OQ}\cdot\overrightarrow{OA}=\dfrac{5(5r-1)+5(5r-1)+10(10r+2)}{r+1}=\dfrac{150r+10}{r+1}$.
$\overrightarrow{OP}\times\overrightarrow{OA}=\dfrac{20r}{r+1}(-\hat{i}+\hat{j})$, $|\overrightarrow{OP}\times\overrightarrow{OA}|^2=\dfrac{800r^2}{(r+1)^2}$.
$\dfrac{150r+10}{r+1}-\dfrac{160r^2}{(r+1)^2}=10$.
Multiply by $(r+1)^2$: $(150r+10)(r+1)-160r^2=10(r+1)^2$.
$150r^2+160r+10-160r^2=10r^2+20r+10 \Rightarrow -20r^2+140r=0 \Rightarrow r(r-7)=0$.
Since $r>0$: $r=7$.
Correct Answer: 4