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 $b|\overrightarrow{R}|$. The general relation among vectors $\overrightarrow{P},\overrightarrow{Q}$ and $\overrightarrow{S}$ is

• A. $\overrightarrow{S}=(1-b)\overrightarrow{P}+b^2\overrightarrow{Q}$
• B. $\overrightarrow{S}=(1-b^2)\overrightarrow{P}+b\overrightarrow{Q}$
• C. $\overrightarrow{S}=(1-b)\overrightarrow{P}+b\overrightarrow{Q}$
• D. $\overrightarrow{S}=(b-1)\overrightarrow{P}+b\overrightarrow{Q}$

Answer 1

The correct option is (C) $\overrightarrow{S}=(1-b)\overrightarrow{P}+b\overrightarrow{Q}$

Step 1, Given data

Given figure is

Step 2,

From the triangular law of vector addition, we get

$OP+PS=OS$

Putting values

$\therefore$ $\overrightarrow{P}+b|\overrightarrow{R}|\frac{\overrightarrow{R}}{|\overrightarrow{R}|}=\overrightarrow{S}$

Or we can write

$\overrightarrow{P}+b\overrightarrow{R}=\overrightarrow{S}$

But $\overrightarrow{R}=\overrightarrow{Q}-\overrightarrow{P}$ (Given)

So,

$\overrightarrow{P}+b(\overrightarrow{Q}-\overrightarrow{P})=\overrightarrow{S}$

$⟹$ $\overrightarrow{S}=(1-b)\overrightarrow{P}+b\overrightarrow{Q}$

Hence the value of $⟹$ $\overrightarrow{S}=(1-b)\overrightarrow{P}+b\overrightarrow{Q}$