If $\theta$ is the interior angle of a regular pentagon,then $|(\sin \theta) \hat{i}+(\cos \theta) \hat{j}+(\tan \theta) \hat{k}|=$

Let $\vec{i}+\vec{j}+\vec{k}$,$a_1 \vec{i}+b_1 \vec{j}+c_1 \vec{k}$,$a_2 \vec{i}+b_2 \vec{j}+c_2 \vec{k}$,and $a_3 \vec{i}+b_3 \vec{j}+c_3 \vec{k}$ be the position vectors of the points $A, B, C, D$ respectively. The position vector of the centroid of the triangular face $BCD$ is $\frac{2}{3}(\vec{i}+\vec{j}+\vec{k})$. If $\alpha \vec{i}+\beta \vec{j}+\gamma \vec{k}$ is the position vector of the centroid of the tetrahedron $ABCD$,then find the value of $2 \alpha+\beta+\gamma$.

If the vectors $\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}$ and $\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}$ are collinear,then the values of $p$ and $q$ are:

If $ABCDEF$ is a regular hexagon,then $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = $

Represent graphically a displacement of $40 \, km$,$30^{\circ}$ east of north.

Find the direction cosines of the vector $\hat{i}+2 \hat{j}+3 \hat{k}$.