Name: Scalar and Vector Triple Product
Uploaded: 2019-07-19
Duration: 12 min 33 s
Description: This video explains the triple scalar product and its use in finding the volume of a parallelepiped and determining if three vectors are coplanar.

Question

If $\vec{a}, \vec{b}, \vec{c} and {\vec{a}}^{'}, {\vec{b}}^{'}, {\vec{c}}^{'}$ are the reciprocal system of vectors, then prove that ${\vec{a}}^{'} \times {\vec{b}}^{'} + {\vec{b}}^{'} \times {\vec{c}}^{'} + {\vec{c}}^{'} \times {\vec{a}}^{'} = \frac{\vec{a} + \vec{b} + \vec{c}}{[\vec{a} \vec{b} \vec{c}]}$

Accepted Answer

Given, ${\vec{a}}^{'}, {\vec{b}}^{'} and {\vec{c}}^{'}$ are reciprocal of $\vec{a}, \vec{b}, \vec{c}$ .

To prove that:

${\vec{a}}^{'} \times {\vec{b}}^{'} + {\vec{b}}^{'} \times {\vec{c}}^{'} + {\vec{c}}^{'} \times {\vec{a}}^{'} = \frac{\vec{a} + \vec{b} + \vec{c}}{[\vec{a} \vec{b} \vec{c}]}$

$LHS :$

' $= {\vec{a}}^{'} \times {\vec{b}}^{'} + {\vec{b}}^{'} \times {\vec{c}}^{'} + {\vec{c}}^{'} \times {\vec{a}}^{'}$

$= \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]} \times {\vec{b}}^{'} + \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]} \times {\vec{c}}^{'} + \frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]} \times {\vec{a}}^{'}$

$= \frac{1}{[\vec{a} \vec{b} \vec{c}]} \{(\vec{b} \cdot {\vec{b}}^{'}) \vec{c} - ({\vec{b}}^{'} \cdot \vec{c}) \vec{b} + (\vec{c} \cdot {\vec{c}}^{'}) \vec{a} - ({\vec{c}}^{'} \cdot \vec{a}) \vec{c} + ({\vec{a}}^{'} \cdot \vec{a}) \vec{b} - ({\vec{a}}^{'} \cdot \vec{b}) \vec{a}\}$

$= \frac{1}{[\vec{a} \vec{b} \vec{c}]} (\vec{a} + \vec{b} + \vec{c})$

$= \frac{(\vec{a} + \vec{b} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]} = RHS$

Important Questions on Product of Vectors

Learn and Prepare for any exam you want !

If a→, b→, c→ and a→', b→', c→' are the reciprocal system of vectors, then prove that a→'×b→'+b→'×c→'+c→'×a→'=a→+b→+c→[a→ b→ c→]

Important Questions on Product of Vectors

Learn and Prepare for any exam you want !