Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is:

If $A, B, C, D$ are any four points in space, then $|\overrightarrow{AB} \times \overrightarrow{CD} + \overrightarrow{BC} \times \overrightarrow{AD} + \overrightarrow{CA} \times \overrightarrow{BD}|$ is equal to (where $\Delta$ denotes the area of $\Delta ABC$) (in $\Delta$)

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

The unit vector perpendicular to the vectors $6i + 2j + 3k$ and $3i - 6j - 2k$ is

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

If non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other,then the solution of the equation $\vec{r} \times \vec{a} = \vec{b}$ is: