The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is ....... $^o$
$120$
$150$
$135$
None of these
The position vectors of points $A, B, C$ and $D$ are $\vec A = 3\hat i + 4\hat j + 5\hat k,\,\vec B = 4\hat i + 5\hat j + 6\hat k,\,\vec C = 7\hat i + 9\hat j + 3\hat k$ and $\vec D = 4\hat i + 6\hat j$ then the displacement vectors $\overrightarrow {AB} $ and $\overrightarrow {CD} $ are
The value of the sum of two vectors $\overrightarrow A $ and $\overrightarrow B $ with $\theta $ as the angle between them is
An object of $m\, kg$ with speed of $v\, m/s$ strikes a wall at an angle $\theta$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be
The coordinates of a moving particle at any time $t$ are given by $x = a\, t^2$ and $y = b\, t^2$. The speed of the particle is
Two vectors $\dot{A}$ and $\dot{B}$ are defined as $\dot{A}=a \hat{i}$ and $\overrightarrow{\mathrm{B}}=\mathrm{a}(\cos \omega t \hat{\mathrm{i}}+\sin \omega t \hat{j}$ ), where a is a constant and $\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1}$. If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $t=\tau$ for the first time, the value of $\tau$, in, seconds, is. . . . . .