Exam

JEE Main

Year

2025

Date

23 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Three Dimensional Geometry

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let the point A divide the line segment joining the points...

Let the point A divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $\mathrm{r}: 1(\mathrm{r}\gt0)$. If O is the origin and $(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=10$, then the value of r is :

Step-by-step solution

$$\begin{aligned} & A=\left(\frac{5 r-1}{r+1}, \frac{5 r-1}{r+1}, \frac{10 r+2}{r+1}\right) \\ & (\overrightarrow{O Q} \cdot \overrightarrow{O A})-\frac{1}{5}|\overrightarrow{O P} \times \overrightarrow{O A}|^2=10 \\ & \overrightarrow{O Q}=5 \hat{i}+5 \hat{j}+10 \hat{k} \\ & \overrightarrow{O A}=\frac{5 r-1}{r+1} \hat{i}+\frac{5 r-1}{r+1} \hat{j}+\frac{10 r+2}{r+1} \hat{k} \\ & \overrightarrow{O P}=-\hat{i}-\hat{j}+2 \hat{k} \\ & \left.\overrightarrow{O P} \times \overrightarrow{O A}=\frac{1}{r+1} \right\rvert\, \frac{5 r-1}{5} \quad 5 r-1 \quad 10 r+2 \\ & =\frac{1}{r+1}(\hat{i}(20 r)-\hat{j}(20 r)) \\ & =5\left(\frac{5 r-1}{r+1}\right)+5\left(\frac{5 r-1}{r+1}\right)+10\left(\frac{10 r+2}{r+1}\right) \\ & \quad-\frac{1}{5}\left(\frac{2 \times 400 r^2}{(r+1)^2}\right)=10 \\ & \frac{150 r+10}{r+1}-\frac{1}{5}\left(\frac{2 \times 400 r^2}{(r+1)^2}\right)=10 \\ & (150 r+10)(r+1)-160 r^2=10(r+1)^2 \\ & (15 r+1)(r+1)-16 r^2=(r+1)^2 \\ & 15 r^2+16 r+1-16 r^2=r^2+2 r+1 \\ & -2 r^2+14 r=0\end{aligned}$$r=0,7$