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

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

Explore More

Similar Questions

If $a = 3i - 2j + k$,$b = 2i - 4j - 3k$ and $c = -i + 2j + 2k$,then $a + b + c$ is

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are position vectors of $4$ points such that $2 \vec{a}+3 \vec{b}+5 \vec{c}-10 \vec{d}=\vec{0}$,then the ratio in which the line joining $\vec{c}$ and $\vec{d}$ divides the line segment joining $\vec{a}$ and $\vec{b}$ is

Let $\hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-2 \hat{k}$ be the position vectors of points $A$ and $B$ respectively. If $C$ is a point on the line joining $A$ and $B$ such that $BC=10$,then the position vector of $C$ can be

If $(\alpha, \beta, \gamma)$ is a triad of real numbers satisfying $\hat{i}-2 \hat{j}+5 \hat{k}=\alpha(\hat{i}+\hat{j}+\hat{k})+\beta(\hat{i}+2 \hat{j}+3 \hat{k})+\gamma(2 \hat{i}-\hat{j}+\hat{k}),$ then $\alpha^2-\beta^2+\gamma^2=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo