In the given figure (a square),identify the co-initial vectors.

If the vector $\bar{i}-7 \bar{j}+2 \bar{k}$ is along the internal bisector of the angle between the vectors $\bar{a}$ and $-2 \bar{i}-\bar{j}+2 \bar{k}$ and the unit vector along $\bar{a}$ is $x \bar{i}+y \bar{j}+z \bar{k}$ then $x=$

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$,then $3 \lambda$ is equal to :

In a triangle $ABC$,if $\overline{BC} = \bar{i} - 2\bar{j} + 2\bar{k}$ and $\overline{CA} = 6\bar{i} + 3\bar{j} - 2\bar{k}$,then the perimeter of the triangle is

Let $u$ and $v$ be two vectors. Then $|u-v|=||u|-|v||$ if and only if

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