Let $\vec{\alpha}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$. Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$,then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is

If $\vec{a}$ is a nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$,$\hat{i}+2 \hat{j}-2 \hat{k}$,and $\hat{k}$ are equal,then a unit vector along $\vec{a}$ is:

If $ABCDEF$ is a regular hexagon,the length of whose side is $a$,then $\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = $

$A$ vector $a$ makes equal acute angles with the coordinate axes. Then the projection of vector $b = 5\hat{i} + 7\hat{j} - \hat{k}$ on $a$ is

If $a, b, c$ are three non-zero,non-coplanar vectors and $b_1 = b - \frac{b \cdot a}{|a|^2} a$,$b_2 = b + \frac{b \cdot a}{|a|^2} a$,$c_2 = c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1$,$c_3 = c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_2}{|b_2|^2} b_2$,and $c_4 = a - \frac{c \cdot a}{|a|^2} a$. Then which of the following is a set of mutually orthogonal vectors?

Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$,$\vec{b}=6 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{c}=3 \hat{i}-4 \hat{j}-12 \hat{k}$ be three vectors. If $\vec{p}$ is the projection of $\vec{b}$ on $\vec{a}$ and $\vec{q}$ is the projection of $\vec{c}$ on $\vec{a}$,then $13 \vec{p}=$ (in $vec{q}$)