The area of the triangle whose vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ is . . . . . . sq. units.

If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them,then for what value of $\theta$ is $2\hat{u} \times 3\hat{v}$ a unit vector?

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.

$\vec{a}, \vec{b}, \vec{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\vec{a}, \vec{b})=(\vec{b}, \vec{c})=(\vec{c}, \vec{a})=\frac{\pi}{3}$. If $\vec{x}=\vec{a} \times(\vec{b} \times \vec{c})$ and $\vec{y}=\vec{b} \times(\vec{c} \times \vec{a})$, then

Let $a, b$ and $c$ be unit vectors such that $a \cdot b = 0 = a \cdot c$ and the acute angle between $b$ and $c$ is $\frac{\pi}{3}$,then $|a \times b - a \times c|$ is equal to

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $XY$-plane. Then,the maximum value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is: