Three forces of magnitudes $1, 2, 3 \text{ dynes}$ meet at a point and act along the diagonals of three adjacent faces of a cube. The resultant force is ............ $\text{dyne}$.

If the position vectors of the points $A, B, C, D$ are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 5\hat{j}$,$\vec{c} = 3\hat{i} + 2\hat{j} - 3\hat{k}$,and $\vec{d} = \hat{i} - 6\hat{j} - \hat{k}$,then the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is:

$ABCD$ is a parallelogram and $P$ is the mid-point of the side $AD$. The line $BP$ meets the diagonal $AC$ in $Q$. Then,the ratio of $AQ:QC$ is equal to

$A$ vector of magnitude $14$ lies in the $xy-$ plane and makes an angle of $60^\circ$ with the $x-$ axis. The components of the vector in the direction of the $x-$ axis and $y-$ axis are:

The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

The points whose position vectors are $2\hat{i}+3\hat{j}+4\hat{k}$,$3\hat{i}+4\hat{j}+2\hat{k}$,and $4\hat{i}+2\hat{j}+3\hat{k}$ are the vertices of