Question
The three concurrent edges of a parallelepiped represent the vectors $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ such that $$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = \lambda .$$ Then the volume of the parallelepiped whose three concurrent edges are the three concurrent diagonals of three faces of the given parallelepiped is :
A.
$$2\lambda $$
B.
$$3\lambda $$
C.
$$\lambda $$
D.
none of these
Answer :
$$2\lambda $$
Solution :
The new edges of the new parallelepiped will be $$\left( {\overrightarrow a + \overrightarrow b } \right),\,\left( {\overrightarrow b + \overrightarrow c } \right),\,\left( {\overrightarrow c + \overrightarrow a } \right)$$
Hence,
The parallelepiped thus formed will have a volume
$$\eqalign{
& \left[ {\left( {\overrightarrow a + \overrightarrow b } \right)\,\,\,\left( {\overrightarrow b + \overrightarrow c } \right)\,\,\,\left( {\overrightarrow c + \overrightarrow a } \right)} \right] \cr
& = 2\left[ {\overrightarrow a \,\,\,\overrightarrow b \,\,\,\overrightarrow c } \right] \cr
& = 2\lambda \cr} $$
Hence,
The volume of the new parallelepiped is equal to $$2\lambda $$.