$A$ vector $\vec{A}$ points vertically upward and $\vec{B}$ points towards north. The vector product $\vec{A} \times \vec{B}$ is

If $\vec{a}$ and $\vec{b}$ make an angle $\cos^{-1}\left(\frac{5}{9}\right)$ with each other,then $|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$ for $|\vec{a}|=n|\vec{b}|$. The integer value of $n$ is . . . . . . .

The area of the parallelogram whose sides are represented by the vectors $\hat j + 3\hat k$ and $\hat i + 2\hat j - \hat k$ is

The area of the parallelogram determined by vectors $A = 2\hat{i} + \hat{j} - 3\hat{k}$ and $B = 12\hat{j} - 2\hat{k}$ is approximately:

Show that the area of the triangle formed by the vectors $\vec{a}$ and $\vec{b}$ is one-half of the magnitude of $\vec{a} \times \vec{b}$.

If $F_1$ and $F_2$ are two vectors of equal magnitudes $F$ such that $|F_1 \cdot F_2| = |F_1 \times F_2|$,then $|F_1 + F_2|$ is equal to