Question

In a two dimensional motion of a particle, the particle moves from point A, position vector r1 to point B position vector r2. If the magnitude of these vector are respectively r1=3 and r2=4 and the angle they make with the x- axis are theta 1 =75° and theta 2=15° respectively, then magnitude of the displacement vector is -

Answer 1

displacement is the minimum path covered by particle or you can say that it is the difference of final position to initial position.

here,

initial position vector of particle is $\bf{r_1}$

final position vector of particle is $\bf{r_2}$

so, displacement = $\bf{r_2-r_1}$

magnitude of displacement = $\sqrt{|r_1|^2+|r_2|^2-2|r_1||r_2|cos\theta}$

here $\theta$ is angle between them.

angle between $r_1$ and $r_2$ = $\theta_1-\theta_2$

= 75° - 15° = 60°

and also given, $|r_1|=3,|r_2|=4$

so, magnitude of displacement= √{3² + 4² - 2(3)(4)cos60°}

= √{9 + 16 - 24 × 1/2}

= √13 unit