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Question

Show that the four points P, Q, R, S with position vectors p, q, r, s respectively such that 5p - 2q + 6r - 9s = 0, are coplaner. Also, find the position vector of the point of intersection of the line segment PR and QS.

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Solution

5p2q+6r9s=0
5p5q+3q3r+9r9s=0
5(pq)+3(qr)+9(rs)=0
5QP+3RQ+9SR=0
IfP,Q,Rand S are coplanar,
We know P,Q,R are coplanar,
3 points always lie in the same plane and hence we have to only check if S lies in this plane or not.
If S lies in the same plane then
SR.(QP×QR)=0(1)
5QP3RQ9.(QP×QR)
19[5QP.(QP×QR)3RQ.(QP×(RQ))]
19[5QP.(QP×QR)+3RQ.(QP×RQ)]
QP.(QP×QR)=0
and RQ.(QP×RQ)=0
( Dot product of perpendicular vector =0)
(1)=SR.(QP×QR)=0
P,Q,R and S are coplanar
Now, 5p+6r=2q+9s
Dividing by 11, we get
5p+6r11=2q+9s11
By section formula, we can say that point of intersection of PRandQS divides PR segment in the ratio 9:2
Point of intersection is 5p+6r11 or 2q+9s11

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