The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $1$ and $2$ respectively,given that $\vec{a} \cdot \vec{b} = 1$.

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality:

If $\bar{a}$ and $\bar{b}$ are two unit vectors such that $\bar{a}+2 \bar{b}$ and $5 \bar{a}-4 \bar{b}$ are perpendicular to each other,then the angle between $\bar{a}$ and $\bar{b}$ is

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

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$