Let $$OA$$ $$CB$$ be a parallelogram with $$O$$ at the origin and $$OC$$ a diagonal. Let $$D$$ be the midpoint of $$OA.$$ Using vector methods prove that $$BD$$ and $$CO$$ intersect in the same ratio. Determine this ratio.

If $$A, B, C, D$$ are any four points in space, prove that -
$$\left| {\overrightarrow {AB} \times \overrightarrow {CD} + \overrightarrow {BC} \times \overrightarrow {AD} + \overrightarrow {CA} \times \overrightarrow {BD} } \right| = 4$$ (area of triangle $$ABC$$)

The position vectors of the points $$A, B, C$$ and $$D$$ are $$3\widehat i - 2\widehat j - \widehat k,\,2\widehat i + 3\widehat j - 4\widehat k,\, - \widehat i + \widehat j + 2\widehat k$$ and $$4\widehat i + 5\widehat j + \lambda \widehat k,$$
respectively. If the points $$A, B, C$$ and $$D$$ lie on a plane, find the value of $$\lambda .$$

$${A_1},{A_2},.................{A_n}$$ are the vertices of a regular plane polygon with $$n$$ sides and $$O$$ is its centre. Show that
$$\sum\limits_{i = 1}^{n - 1} {\left( {\overrightarrow {O{A_i}} \times {{\overrightarrow {OA} }_{i + 1}}} \right) = \left( {1 - n} \right)\left( {{{\overrightarrow {OA} }_2} \times {{\overrightarrow {OA} }_1}} \right)} $$