Let → a , → b and → c be three non zero vectors such that → b · → c...

Given,

$\overset{\to}{a} \times \left(\overset{\to}{b} \times \overset{\to}{c}\right) = \dfrac{\overset{\to}{b} - \overset{\to}{c}}{2}$

Using the vector triple cross product we have,

$\overset{\to}{a} \times \left(\overset{\to}{b} \times \overset{\to}{c}\right) = \left(\overset{\to}{a} \cdot \overset{\to}{c}\right)\overset{\to}{b} - \left(\overset{\to}{a} \cdot \overset{\to}{b}\right)\overset{\to}{c}$

So, $( \overset{\to}{a} \cdot \overset{\to}{c} ) \overset{\to}{b} - ( \overset{\to}{a} \cdot \overset{\to}{b} ) \overset{\to}{c} = \dfrac{\overset{\to}{b} - \overset{\to}{c}}{2}$

On comparing both sides we get,

$\overset{\to}{a} \cdot \overset{\to}{c} = \dfrac{1}{2}, \overset{\to}{a} \cdot \overset{\to}{b} = \dfrac{1}{2}$

$∵ \overset{\to}{b} \cdot \overset{\to}{d} = \overset{\to}{a} \cdot \overset{\to}{b}$

$∴ \overset{\to}{b} \cdot \overset{\to}{d} = \dfrac{1}{2}$

Using vector cross-product formulae we have

$( \overset{\to}{a} \times \overset{\to}{b} ) \cdot ( \overset{\to}{c} \times \overset{\to}{d} ) = \overset{\to}{a} \cdot ( \overset{\to}{b} \times ( \overset{\to}{c} \times \overset{\to}{d} ) )$

$= \overset{\to}{a} \cdot ( ( \overset{\to}{b} \cdot \overset{\to}{d} ) \overset{\to}{c} - ( \overset{\to}{b} \cdot \overset{\to}{c} ) \overset{\to}{d} ) \left(∵ \overset{\to}{b} \cdot \overset{\to}{c} = 0\right)</p><p>$= ( \overset{\to}{a} \cdot \overset{\to}{c} ) ( \overset{\to}{b} \cdot \overset{\to}{d} ) = $\dfrac{1}{2}$ \times $\dfrac{1}{2} = \dfrac{1}{4}</p>$