Given that $\overrightarrow A + \overrightarrow B + \overrightarrow C= 0$ out of three vectors two are equal in magnitude and the magnitude of third vector is $\sqrt 2 $ times that of either of the two having equal magnitude. Then the angles between vectors are given by
$30°, 60°, 90°$
$45°, 45°, 90°$
$45°, 60°, 90°$
$90°, 135°, 135°$
The angle between vector $\vec{Q}$ and the resultant of $(2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}})$ and $(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})$ is:
If $|\,\vec A + \vec B\,|\, = \,|\,\vec A\,| + |\,\vec B\,|$, then angle between $\vec A$ and $\vec B$ will be ....... $^o$
Two forces ${F_1} = 1\,N$ and ${F_2} = 2\,N$ act along the lines $x = 0$ and $y = 0$ respectively. Then the resultant of forces would be
The value of the sum of two vectors $\overrightarrow A $ and $\overrightarrow B $ with $\theta $ as the angle between them is
Three girls skating on a circular ice ground of radius $200 \;m$ start from a point $P$ on the edge of the ground and reach a point $Q$ diametrically opposite to $P$ following different paths as shown in Figure. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?