Let $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{w} = \hat{i} + 3\hat{k}$. If $\vec{u}$ is a unit vector,then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

Consider $\overrightarrow{r}, \overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero vectors such that $\overrightarrow{r} \cdot \overrightarrow{a}=0$,$|\overrightarrow{r} \times \overrightarrow{b}|=|\overrightarrow{r}||\overrightarrow{b}|$,and $|\overrightarrow{r} \times \overrightarrow{c}|=|\overrightarrow{r}||\overrightarrow{c}|$. Then,the scalar triple product $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is:

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,then $x$ equals:

The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors $\hat{a}, \hat{b}, \hat{c}$ such that $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = 1/2$. Then the volume of the parallelepiped is

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

If $\vec{a} = 4\hat{i} - 2\hat{j} + \hat{k}$,$\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}$,and $\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ represent the three coterminous edges of a parallelepiped,find its volume.