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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}] 
eq 0$.Expand the expression $

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$\frac{1}{12}$

Vedclass · Answer

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}] 
eq 0$.Expand the expression $(\vec{a} + \lambda \vec{b}) \cdot [(\vec{b} + 3\vec{c}) 	imes (\vec{c} - 4\vec{a})] = 0$.First,compute the cross product: $(\vec{b} + 3\vec{c}) 	imes (\vec{c} - 4\vec{a}) = \vec{b} 	imes \vec{c} - 4(\vec{b} 	imes \vec{a}) + 3(\vec{c} 	imes \vec{c}) - 12(\vec{c} 	imes \vec{a})$.Since $\vec{c} 	imes \vec{c} = 0$,this simplifies to $\vec{b} 	imes \vec{c} + 4(\vec{a} 	imes \vec{b}) - 12(\vec{c} 	imes \vec{a})$.Now,take the dot product with $(\vec{a} + \lambda \vec{b})$:$(\vec{a} + \lambda \vec{b}) \cdot [\vec{b} 	imes \vec{c} + 4(\vec{a} 	imes \vec{b}) - 12(\vec{c} 	imes \vec{a})] = 0$.Distributing the dot product:$\vec{a} \cdot (\vec{b} 	imes \vec{c}) + 4\vec{a} \cdot (\vec{a} 	imes \vec{b}) - 12\vec{a} \cdot (\vec{c} 	imes \vec{a}) + \lambda \vec{b} \cdot (\vec{b} 	imes \vec{c}) + 4\lambda \vec{b} \cdot (\vec{a} 	imes \vec{b}) - 12\lambda \vec{b} \cdot (\vec{c} 	imes \vec{a}) = 0$.Using the property that scalar triple products with repeated vectors are zero:$[\vec{a} \, \vec{b} \, \vec{c}] + 0 - 0 + 0 + 0 - 12\lambda [\vec{b} \, \vec{c} \, \vec{a}] = 0$.Since $[\vec{b} \, \vec{c} \, \vec{a}] = [\vec{a} \, \vec{b} \, \vec{c}]$,we have:$(1 - 12\lambda) [\vec{a} \, \vec{b} \, \vec{c}] = 0$.Since $[\vec{a} \, \vec{b} \, \vec{c}] 
eq 0$,we must have $1 - 12\lambda = 0$,which gives $\lambda = \frac{1}{12}$.

If $\vec{a}, \vec{b}, \vec{c}$ are non-zero and non-coplanar vectors such that $(\vec{a} + \lambda \vec{b}) \cdot [(\vec{b} + 3\vec{c}) \times (\vec{c} - 4\vec{a})] = 0$,then $\lambda$ is equal to

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