Question
Three vectors $$\overrightarrow P ,$$ $$\overrightarrow Q $$ and $$\overrightarrow R $$ are shown in the figure. Let $$S$$ be any point on the vector $$\overrightarrow R .$$ The distance between the points $$P$$ and $$S$$ is $$b\left| {\vec R} \right|.$$ The general relation among vectors $$\overrightarrow P ,$$ $$\overrightarrow Q $$ and $$\overrightarrow S $$ is-
Three vectors $$\overrightarrow P ,$$ $$\overrightarrow Q $$ and $$\overrightarrow R $$ are shown in the figure. Let $$S$$ be any point on the vector $$\overrightarrow R .$$ The distance between the points $$P$$ and $$S$$ is $$b\left| {\vec R} \right|.$$ The general relation among vectors $$\overrightarrow P ,$$ $$\overrightarrow Q $$ and $$\overrightarrow S $$ is-
A.
$$\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + b\overrightarrow Q $$
B.
$$\overrightarrow S = \left( {b - 1} \right)\overrightarrow P + b\overrightarrow Q $$
C.
$$\overrightarrow S = \left( {1 - {b^2}} \right)\overrightarrow P + b\overrightarrow Q $$
D.
$$\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + {b^2}\overrightarrow Q $$
Answer :
$$\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + b\overrightarrow Q $$
Solution :
$$\eqalign{ & {\text{Here }}\vec P + b\vec R = \vec S \cr & \therefore \vec R = \frac{{\vec S - \vec P}}{b} \cr & {\text{Also}}\,\vec R = \vec Q - \vec P \cr & \therefore \frac{{\vec S - \vec P}}{b} = \vec Q - \vec P \cr & \therefore \vec S - \vec P = b\vec Q - b\vec P \cr & \therefore \vec S = b\vec Q + \left( {1 - b} \right)\vec P \cr} $$
$$\eqalign{ & {\text{Here }}\vec P + b\vec R = \vec S \cr & \therefore \vec R = \frac{{\vec S - \vec P}}{b} \cr & {\text{Also}}\,\vec R = \vec Q - \vec P \cr & \therefore \frac{{\vec S - \vec P}}{b} = \vec Q - \vec P \cr & \therefore \vec S - \vec P = b\vec Q - b\vec P \cr & \therefore \vec S = b\vec Q + \left( {1 - b} \right)\vec P \cr} $$
