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Question:

Three vectors →P, →Q and →R are shown in the figure. Let S be any point on the vector →R. The distance between the points P and S is b|→R|. The general relation among vectors →P, →Q and →S is

→S=(1−b)→P+b²→Q

→S=(b𕒵)→P+b→Q

→S=(1−b²)→P+b→Q

→S=(1−b)→P+b→Q

Solution:

The correct option is D →S=(1−b)→P+b→Q
From triangular law of vector addition, we get
OP + PS = OS
∴ →P + b|→R| = →S
→P + b→R = →S
But →R = →Q − →P (Given)
→P + b(→Q − →P) = →S
⇒ →S = (1−b)→P + b→Q