If C is the mid point of AB and P is any point outside AB, then
A
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB+¯¯¯¯¯¯¯¯PC=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB+2¯¯¯¯¯¯¯¯PC=¯0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=¯¯¯¯¯¯¯¯PC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=2¯¯¯¯¯¯¯¯PC
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=2¯¯¯¯¯¯¯¯PC Let →p,→a,→b,→c be the position vectors of points P,A,B,C respectively. L.H.S=¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=→p−→a+→p−→b=2(→p−→a+→b2)=2→PC