The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.

Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$,$\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$,$\vec{a} \cdot \vec{c}=7$,$2 \vec{b} \cdot \vec{c}+43=0$,and $\vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

Let $a, b$ and $c$ be three coplanar unit vectors. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{3}{26} i - \frac{2}{13} j + \frac{6}{13} k$ and the angle between $a$ and $b$ is $30^\circ$, then $c$ is equal to...

If $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$,then the vector $\overline{r}$ satisfying $\overline{a} \times \overline{r} = \overline{b}$ and $\overline{a} \cdot \overline{r} = 3$ is

Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

$a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d)$ is equal to