Three vectors vec{P}, vec{Q} and vec{R} are s | vedclass

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(A) From the figure,the vector $\vec{R}$ is given by $\vec{R} = \vec{Q} - \vec{P}$.Point $S$ lies on the line segment $PQ$ such that the distance $PS

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) From the figure,the vector $\vec{R}$ is given by $\vec{R} = \vec{Q} - \vec{P}$.Point $S$ lies on the line segment $PQ$ such that the distance $PS = b|\vec{R}|$.This implies that the vector $\vec{PS} = b\vec{R} = b(\vec{Q} - \vec{P})$.Using the triangle law of vector addition in $	riangle OPS$,we have $\vec{S} = \vec{P} + \vec{PS}$.Substituting the expression for $\vec{PS}$,we get $\vec{S} = \vec{P} + b(\vec{Q} - \vec{P})$.Simplifying this,we obtain $\vec{S} = \vec{P} + b\vec{Q} - b\vec{P} = (1-b)\vec{P} + b\vec{Q}$.

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

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