If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=\hat{j}-\hat{k}$,find a vector $\bar{c}$ such that $\bar{a} \times \bar{c}=\bar{b}$ and $\bar{a} \cdot \bar{c}=3$.

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors of unit modulus. If $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v}=\vec{a} \times \vec{b}$,then $|\vec{v}|=$

Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

Let $\overrightarrow{OA}=2 \overrightarrow{a}$,$\overrightarrow{OB}=6 \overrightarrow{a}+5 \overrightarrow{b}$ and $\overrightarrow{OC}=3 \overrightarrow{b}$,where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units,then the area (in sq. units) of the quadrilateral $OABC$ is equal to :

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0},$ then $\vec{a} \times \vec{b}=\overrightarrow{0} .$ Is the converse true? Justify your answer with an example.

Let $m$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. Let $n$ be another vector of magnitude $2\sqrt{6}$ and perpendicular to the vectors $2\hat{i}-\hat{j}$ and $\hat{j}+2\hat{k}$. The area (in sq. units) of the triangle formed with $m$ and $n$ as sides is