Verified Solution • 2025–26 Session moderate

QJEEM25Ma2623: JEE Main 2025 Physics - Class 11 JEEM25Ma2623 Solution: Let the point A divide the line segment joining the points and…

Let the point A divide the line segment joining the points P(−1,−1,2)\mathrm{P}(-1,-1,2) and Q(5,5,10)\mathrm{Q}(5,5,10) internally in the ratio r: 1(r>0) . If O is the origin and (OQ→⋅OA→)−15∣OP→×OA→∣2=10(\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 :

Multiple Choice Options

  • A 7\sqrt7
  • B 14
  • C 7
  • D 3

Step-by-Step Verified Solution

amp;amp;A≡(5r−1r+1,5r−1r+1,10r+2r+1)amp;(OQ→⋅OA→)−∣OP→×OA→∣25=10.... (1)amp;OQ→⋅OA→=10r+1(15r+1)amp;∣OP→×OA→∣2=r2(r+1)2(800)amp; so by equation (1) amp;amp;10r+1(15r+1)−15r2(800)(r+1)2=10amp;2r2−14r=0amp;r=7,r≠0\begin{aligned}&\begin{aligned}& \mathrm{A} \equiv\left(\frac{5 \mathrm{r}-1}{\mathrm{r}+1}, \frac{5 \mathrm{r}-1}{\mathrm{r}+1}, \frac{10 \mathrm{r}+2}{\mathrm{r}+1}\right) \\& (\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2}{5}=10 \quad\text{.... (1)}\\& \overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}}=\frac{10}{\mathrm{r}+1}(15 \mathrm{r}+1) \\& |\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=\frac{\mathrm{r}^2}{(\mathrm{r}+1)^2}(800)\end{aligned}\\&\text { so by equation (1) }\\&\begin{aligned}& \frac{10}{r+1}(15 r+1)-\frac{1}{5} \frac{r^2(800)}{(r+1)^2}=10 \\& 2 r^2-14 r=0 \\& r=7, r \neq 0\end{aligned}\end{aligned}

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