The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i} - 6\hat{j}$ and $3\hat{i} + 4\hat{j} - 12\hat{k}$ is:

If the vertices of $\Delta ABC$ are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$,then the area of the $\Delta ABC$ (in sq. units) is

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$,$\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$,then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to:

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ be two vectors such that $|\vec{a}|=1$,$\vec{a} \cdot \vec{b}=2$,and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$,then the angle between $\vec{b}$ and $\vec{c}$ is equal to:

Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b},$ where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}.$

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{a} \cdot \vec{b} = 1$,and $\vec{a} \times \vec{b} = \hat{j} - \hat{k}$,then $\vec{b} = \dots$