Two particles $A$ and $B$ move with constant velocities $\vec{v}_1$ and $\vec{v}_2$. At the initial moment,their position vectors are $\vec{r}_1$ and $\vec{r}_2$ respectively. The condition for their collision is

  • A
    $\vec{r}_1 - \vec{r}_2 = \vec{v}_1 - \vec{v}_2$
  • B
    $\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$
  • C
    $\vec{r}_1 \cdot \vec{v}_1 = \vec{r}_2 \cdot \vec{v}_2$
  • D
    $\vec{r}_1 \times \vec{v}_1 = \vec{r}_2 \times \vec{v}_2$

Explore More

Similar Questions

$A$ linear charge having linear charge density $\lambda$ penetrates a cube diagonally and then it penetrates a sphere diametrically as shown. What will be the ratio of flux coming out of the cube and the sphere?

If $a = \hat{i} + \hat{j} + t \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$,then the values of $t$ for which $(a+b)$ and $(a-b)$ are perpendicular are:

If $\Delta = a^2 - (b - c)^2$ is the area of the $\triangle ABC$,then $\tan A$ is equal to

The number of radial nodes of $3s$ and $2p$ orbitals are respectively

At a certain place,the angle of dip is $60^{\circ}$ and the horizontal component of the earth's magnetic field $(B_H)$ is $0.8 \times 10^{-4} ~T$. The earth's overall magnetic field is

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