If the vectors $\vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$ represent two adjacent sides of a parallelogram $ABCD$ and $\theta$ is the angle between its diagonals $\vec{AC}$ and $\vec{BD}$,then $\tan \theta =$

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}+\bar{b}|=\sqrt{29}$ and $\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}$,then a possible value of $(\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

If the vector $a = 3\hat{j} + 4\hat{k}$ is the sum of two vectors $a_1$ and $a_2$,where vector $a_1$ is parallel to $b = \hat{i} + \hat{j}$ and vector $a_2$ is perpendicular to $b$,then $a_1 =$

Let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$ of a triangle $ABC$ respectively. If $O$ is an interior point of the triangle $ABC$ such that $\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=\overrightarrow{0}$,then the area (in sq. units) of the triangle $ODE$ is

Let $ABCD$ be a parallelogram where $\overrightarrow{AB} = \overrightarrow{a}$,$\overrightarrow{AD} = \overrightarrow{b}$,$|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and $|\overrightarrow{a} \times \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$,where $\overrightarrow{a} \cdot \overrightarrow{b} > 0$. Then the area of this parallelogram is (in square units):

Let $\overline{u}, \overline{v}, \overline{w}$ be vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to that of $\overline{w}$ along $\overline{u}$,and the vectors $\overline{v}$ and $\overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|$ equals: