Three vectors $\vec{P}, \vec{Q}$ and $\vec{R}$ are shown in the figure. Let $S$ be any point on the vector $\vec{R}$. The distance between the point $P$ and $S$ is $b|\vec{R}|$. The general relation among vectors $\vec{P}, \vec{Q}$ and $\vec{S}$ is

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

Find the values of $x, y$ and $z$ so that the vectors $\vec{a} = x \hat{i} + 2 \hat{j} + z \hat{k}$ and $\vec{b} = 2 \hat{i} + y \hat{j} + \hat{k}$ are equal.

$(3, 0, 2)$ and $(0, 2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them. If $|\cos \theta| = \frac{6}{13}$,then $k =$

$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is: