$A$ unit vector which is perpendicular to $i + 2j - 2k$ and $-i + 2j + 2k$ is

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:

Let $\vec{a} = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k}$. The projection of the sum of the vectors $\vec{a}$ and $\vec{b}$ on the vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is:

Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$. If $\vec{p}=\vec{a} \times \vec{b}$,then the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$ and $|\vec{p} \times \vec{c}|=3$. Thus,$\vec{a} \cdot \vec{c}$ is equal to:

$A$ vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$,and the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2\hat{j}+2\hat{k}$ is

If the vectors $\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j}$ represent the diagonals of a parallelogram,then its area will be: