$100$ coplanar forces,each equal to $10 \ N$,act on a body. Each force makes an angle of $\pi/50$ radians with the preceding force. What is the resultant of the forces in $N$?

Two particles are located at an equal distance from the origin. The position vectors of these are represented by $\overrightarrow{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k}$ and $\overrightarrow{B} = 2\hat{i} - 2\hat{j} + 4p\hat{k}$,respectively. If both vectors are at a right angle to each other,the value of $n^{-1}$ is . . . . . . .

$A$ vector $\vec{A}$ having magnitude $6$ units is added to vector $\vec{B}$,which is along the $x$-axis. The resultant of $\vec{A}$ and $\vec{B}$ is along the $y$-axis. If the magnitude of the resultant of $\vec{A}$ and $\vec{B}$ is three times that of $\vec{B}$,then the magnitude of $\vec{B}$ is:

The three vectors $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}$,$\vec{B}=\hat{i}-3 \hat{j}+5 \hat{k}$ and $\vec{C}=2 \hat{i}-\hat{j}+4 \hat{k}$ will form

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 ....... $^o$

If $\overrightarrow{A}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\overrightarrow{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k}$ and $\overrightarrow{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}$ form a right-angled triangle,then which of the following is satisfied?