Find a vector of magnitude $6,$ which is perpendicular to both the vectors $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.

Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1, 2, \ldots, 10\}$ is

If $3\hat{i} + 4\hat{j}$ and $-5\hat{i} + 7\hat{j}$ are the vector sides of a triangle,then its area is given by

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is a unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then the unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

If $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$,and $a, b, c$ are the moduli of the vectors $\vec{a}, \vec{b}, \vec{c}$ respectively,then:

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$ and $\vec{a} \cdot \vec{c}=3$. If $\vec{b} \times \vec{c}=\vec{d}$,then $|\vec{a} \cdot \vec{d}|$ is equal to: