Let vec{a}, vec{b} and vec{c} be three non-ze | vedclass

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(D) Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(D) Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Given $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$,we have $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{c}$.Comparing the coefficients of $\vec{b}$ and $\vec{c}$,we get $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = \frac{1}{2}$.Given $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$,so $\vec{b} \cdot \vec{d} = \frac{1}{2}$.Now,consider the scalar triple product $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = \vec{a} \cdot (\vec{b} 	imes (\vec{c} 	imes \vec{d}))$.Using the vector triple product formula $\vec{b} 	imes (\vec{c} 	imes \vec{d}) = (\vec{b} \cdot \vec{d})\vec{c} - (\vec{b} \cdot \vec{c})\vec{d}$.Since $\vec{b} \cdot \vec{c} = 0$,this simplifies to $(\vec{b} \cdot \vec{d})\vec{c}$.Thus,$(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = \vec{a} \cdot ((\vec{b} \cdot \vec{d})\vec{c}) = (\vec{b} \cdot \vec{d})(\vec{a} \cdot \vec{c})$.Substituting the values,we get $\frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.

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