Question

Prove that \[\left( {\overrightarrow p - \overrightarrow q } \right).\left[ {\left( {\overrightarrow q - \overrightarrow r } \right) \times \left( {\overrightarrow r - \overrightarrow p } \right)} \right] = 0\]

Answer 1

Hint:First we will let the given three vectors to be \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] respectively and then add them.
According to the result so obtained we will prove the given equation.
If three vectors are coplanar then the volume of parallelepiped is equal to zero.

Complete step-by-step answer:
First we will consider the left hand side of the equation to be proved:
\[LHS = \left( {\overrightarrow p - \overrightarrow q } \right).\left[ {\left( {\overrightarrow q - \overrightarrow r } \right) \times \left( {\overrightarrow r - \overrightarrow p } \right)} \right]\]
Now let
\[
  \overrightarrow p - \overrightarrow q = \overrightarrow a ....................\left( 1 \right) \\
  \overrightarrow q - \overrightarrow r = \overrightarrow b .....................\left( 2 \right) \\
  \overrightarrow r - \overrightarrow p = \overrightarrow c ......................\left( 3 \right) \\
 \]
Now on adding equations 1, 2 and 3 we get:-
\[
  \overrightarrow a + \overrightarrow b + \overrightarrow c = \left( {\overrightarrow p - \overrightarrow q } \right) + \left( {\overrightarrow q - \overrightarrow r } \right) + \left( {\overrightarrow r - \overrightarrow p } \right) \\
  \overrightarrow a + \overrightarrow b + \overrightarrow c = \left( {\overrightarrow p - \overrightarrow p } \right) + \left( {\overrightarrow q - \overrightarrow q } \right) + \left( {\overrightarrow r - \overrightarrow r } \right) \\
  \overrightarrow a + \overrightarrow b + \overrightarrow c = 0 \\
 \]
Now since the sum of the vectors is zero
Therefore, these vectors are co-planar.
This implies that the volume of parallelepiped having vectors \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] as its edges is zero.
Now since the volume of the parallelepiped is given by the scalar triple product of three vectors
Hence,
\[\left[ {\overrightarrow a {\text{ }}\overrightarrow b {\text{ }}\overrightarrow c } \right] = 0\]
Now since scalar triple product is given by:-
\[\left[ {\overrightarrow a {\text{ }}\overrightarrow b {\text{ }}\overrightarrow c } \right] = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)\]
Therefore,
\[\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right) = 0\]
Now putting back the values from equations 1 ,2 and 3 we get:-
\[\left( {\overrightarrow p - \overrightarrow q } \right).\left[ {\left( {\overrightarrow q - \overrightarrow r } \right) \times \left( {\overrightarrow r - \overrightarrow p } \right)} \right] = 0\]
Therefore,
\[LHS = RHS\]
Hence proved.

Note:When three vectors are co- planar then their scalar triple product as well as the volume of the parallelepiped is always zero.
Scalar triple product can also be written as:
\[
  \left[ {\overrightarrow a {\text{ }}\overrightarrow b {\text{ }}\overrightarrow c } \right] = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right) \\
  \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right) = \left( {\overrightarrow a .\overrightarrow b } \right) \times \left( {\overrightarrow a .\overrightarrow c } \right) \\
 \]