The area of a parallelogram whose adjacent sides are $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = -\hat{j} - 2\hat{k}$ is . . . . . . sq. units.

Let $m$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. Let $n$ be another vector of magnitude $2\sqrt{6}$ and perpendicular to the vectors $2\hat{i}-\hat{j}$ and $\hat{j}+2\hat{k}$. The area (in sq. units) of the triangle formed with $m$ and $n$ as sides is

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...

Let $L_1: \overrightarrow{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R$,$L_2: \overrightarrow{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in R$,and $L_3: \overrightarrow{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in R$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

The locus of the point $P(\vec{r})$ which forms a triangle $ABP$ of area $1$ sq. unit with the fixed points $A(\hat{i})$ and $B(\hat{j})$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors. Suppose $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$. Then $\vec{a}$ is