Find a vector that is coplanar with $\hat{i} + \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\hat{i} + \hat{j} + \hat{k}$.

If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$,then find the angle between $\vec{b}$ and $\vec{c}$.

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. If $\vec c$ is a vector such that $\vec a \cdot \vec c = |\vec c|$,$|\vec c - \vec a| = 2\sqrt 2$,and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$,then $|(\vec a \times \vec b) \times \vec c|$ equals:

If non-zero vectors $a$ and $b$ are perpendicular to each other,then what is the solution for $r \times a = b$?

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$,and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $30^{\circ}$,then the value of $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is equal to