Let the point $A$ divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $r:1\ (r0)$. If $O$ is the origin and $(\overrightarrow{OQ}\cdot\overrightarrow{OA})-\frac15\left|\overrightarrow{OP} imes\overrightarrow{OA} ight|^2=10$, then the value of $r$ is :

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 $(\overrightarrow{OQ}\cdot\overrightarrow{OA})-\frac15\left|\overrightarrow{OP}	imes\overrightarrow{OA}ight|^2=10$, then the value of $r$ is :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\vec{OA}=\frac{r\vec{OQ}+\vec{OP}}{r+1}$. Then $\vec{OQ}\cdot\vec{OA}=\frac{r|Q|^2+P\cdot Q}{r+1}=\frac{150r+(-5-5+20)}{r+1}$ and $\vec{OP}	imes\vec{OA}=\frac r{r+1}\vec{OP}	imes\vec{OQ}$ with $|P	imes Q|^2=|P|^2|Q|^2-(P\cdot Q)^2=6\cdot150-100=800$. The condition gives $\frac{150r+10}{r+1}-\frac{160r^2}{(r+1)^2}=10$, which is satisfied by $r=7$ (check: $\frac{1060}8-\frac{7840}{64}=132.5-122.5=10$). Answer (4).

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?