If vectors vec{a} and vec{b} are not perpendi | vedclass

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(B) Given that $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{d}$.This implies $\vec{b} 	imes (\vec{c} - \vec{d}) = 0$.Since the cross product is zero

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$\vec{c} - \left( \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \right) \vec{b}$

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(B) Given that $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{d}$.This implies $\vec{b} 	imes (\vec{c} - \vec{d}) = 0$.Since the cross product is zero,the vector $(\vec{c} - \vec{d})$ must be parallel to $\vec{b}$.Therefore,we can write $\vec{c} - \vec{d} = \lambda \vec{b}$ for some scalar $\lambda$.This gives $\vec{d} = \vec{c} - \lambda \vec{b}$.Taking the dot product of both sides with $\vec{a}$:$\vec{a} \cdot \vec{d} = \vec{a} \cdot \vec{c} - \lambda (\vec{a} \cdot \vec{b})$.Given $\vec{a} \cdot \vec{d} = 0$,we have $0 = \vec{a} \cdot \vec{c} - \lambda (\vec{a} \cdot \vec{b})$.Thus,$\lambda = \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}$.Substituting $\lambda$ back,we get $\vec{d} = \vec{c} - \left( \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} ight) \vec{b}$.

If vectors $\vec{a}$ and $\vec{b}$ are not perpendicular,and $\vec{c}$ and $\vec{d}$ are two vectors satisfying $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d} = 0$,then the vector $\vec{d}$ is equal to:

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