Verified Solution • 2025–26 Session moderate

QJEEM25Maa6d7: JEE Main 2025 Physics - Class 11 JEEM25Maa6d7 Solution: Let be three points in xy-plane, whose position vector are…

Let A,B,CA, B, C be three points in xy-plane, whose position vector are given by 3i^+j^,i^+3j^\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j} and ai^+(1−a)j^a \hat{i}+(1-a) \hat{j} respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors OA→\overrightarrow{\mathrm{OA}} and OB→\overrightarrow{\mathrm{OB}} is 92\frac{9}{\sqrt{2}} , then the sum of all the possible values of aa is :

Multiple Choice Options

  • A 2
  • B 0
  • C 92\frac{9}{2}
  • D 1

Step-by-Step Verified Solution

 Equation of angle bisector : x−y=0|a(1−a)2|=92⇒a=5 or −4 Sum =5+(−4)=1

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