If $U = [2, -3, 4]$,$X = [0, 2, 3]$,$V = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}$,then $UV + XY = $?

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

If $A=(1,-1,2)$, $B=(3,4,-2)$, $C=(0,3,2)$ and $D=(3,5,6)$, then the angle between the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (in $^{\circ}$)

$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

The value of $\theta \in (0, \pi/2)$ for which vectors $\vec{a} = (\sin \theta)\hat{i} + (\cos \theta)\hat{j}$ and $\vec{b} = \hat{i} - \sqrt{3}\hat{j} + 2\hat{k}$ are perpendicular is

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be: