If vec{a}, vec{b}, vec{c} are non-coplanar ve | vedclass

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(D) Given the scalar triple product: $(\vec{a} - \lambda \vec{b}) \cdot ((\vec{b} - 2\vec{c}) 	imes (\vec{c} + 2\vec{a})) = 0$.First,expand the cross

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$-1/4$

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(D) Given the scalar triple product: $(\vec{a} - \lambda \vec{b}) \cdot ((\vec{b} - 2\vec{c}) 	imes (\vec{c} + 2\vec{a})) = 0$.First,expand the cross product: $(\vec{b} - 2\vec{c}) 	imes (\vec{c} + 2\vec{a}) = \vec{b} 	imes \vec{c} + 2(\vec{b} 	imes \vec{a}) - 2(\vec{c} 	imes \vec{c}) - 4(\vec{c} 	imes \vec{a})$.Since $\vec{c} 	imes \vec{c} = 0$,this simplifies to $\vec{b} 	imes \vec{c} + 2(\vec{b} 	imes \vec{a}) - 4(\vec{c} 	imes \vec{a})$.Now,take the dot product with $(\vec{a} - \lambda \vec{b})$:$(\vec{a} - \lambda \vec{b}) \cdot (\vec{b} 	imes \vec{c} + 2(\vec{b} 	imes \vec{a}) - 4(\vec{c} 	imes \vec{a})) = 0$.Distributing the dot product:$\vec{a} \cdot (\vec{b} 	imes \vec{c}) + 2\vec{a} \cdot (\vec{b} 	imes \vec{a}) - 4\vec{a} \cdot (\vec{c} 	imes \vec{a}) - \lambda \vec{b} \cdot (\vec{b} 	imes \vec{c}) - 2\lambda \vec{b} \cdot (\vec{b} 	imes \vec{a}) + 4\lambda \vec{b} \cdot (\vec{c} 	imes \vec{a}) = 0$.Using the property that the scalar triple product is zero if any two vectors are the same,we have $\vec{a} \cdot (\vec{b} 	imes \vec{a}) = 0$,$\vec{a} \cdot (\vec{c} 	imes \vec{a}) = 0$,$\vec{b} \cdot (\vec{b} 	imes \vec{c}) = 0$,and $\vec{b} \cdot (\vec{b} 	imes \vec{a}) = 0$.This leaves: $\vec{a} \cdot (\vec{b} 	imes \vec{c}) + 4\lambda \vec{b} \cdot (\vec{c} 	imes \vec{a}) = 0$.Since $\vec{b} \cdot (\vec{c} 	imes \vec{a}) = \vec{a} \cdot (\vec{b} 	imes \vec{c})$,we get: $(1 + 4\lambda) [\vec{a} \vec{b} \vec{c}] = 0$.Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$[\vec{a} \vec{b} \vec{c}] 
eq 0$,so $1 + 4\lambda = 0$,which implies $\lambda = -1/4$.

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $(\vec{a} - \lambda \vec{b}) \cdot (\vec{b} - 2\vec{c}) \times (\vec{c} + 2\vec{a}) = 0$,then $\lambda$ is equal to

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