In an octagon $ABCDEFGH$ of equal side, what is the sum of $\overrightarrow{ AB }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ AF }+\overrightarrow{ AG }+\overrightarrow{ AH }$ if, $\overrightarrow{ AO }=2 \hat{ i }+3 \hat{ j }-4 \hat{ k }$
$-16 \hat{i}-24 \hat{j}+32 \hat{k}$
$16 \hat{i}+24 \hat{j}-32 \hat{k}$
$16 \hat{i}+24 \hat{j}+32 \hat{k}$
$16 \hat{i}-24 \hat{j}+32 \hat{k}$
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
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
The vector $\overrightarrow{O A}$ where $O$ is origin is given by $\overrightarrow{O A}=2 \hat{i}+2 \hat{j}$. Now it is rotated by $45^{\circ}$ anticlockwise about $O$. What will be the new vector?
Two vectors $\overrightarrow{{X}}$ and $\overrightarrow{{Y}}$ have equal magnitude. The magnitude of $(\overrightarrow{{X}}-\overrightarrow{{Y}})$ is ${n}$ times the magnitude of $(\overrightarrow{{X}}+\overrightarrow{{Y}})$. The angle between $\overrightarrow{{X}}$ and $\overrightarrow{{Y}}$ is -