The vectors vec{a} and vec{b} are not perpend | vedclass

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(C) Given $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{d}$.Taking the cross product with $\vec{a}$ on both sides: $\vec{a} 	imes (\vec{b} 	imes \ve

Vedclass · Accepted Answer

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

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(C) Given $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{d}$.Taking the cross product with $\vec{a}$ on both sides: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \vec{a} 	imes (\vec{b} 	imes \vec{d})$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$,we get:$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (\vec{a} \cdot \vec{d}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{d}$.Since $\vec{a} \cdot \vec{d} = 0$,the equation simplifies to:$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = 0 - (\vec{a} \cdot \vec{b}) \vec{d}$.Rearranging for $\vec{d}$:$(\vec{a} \cdot \vec{b}) \vec{d} = (\vec{a} \cdot \vec{b}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b}$.Dividing by $(\vec{a} \cdot \vec{b})$ (which is non-zero as $\vec{a}$ and $\vec{b}$ are not perpendicular):$\vec{d} = \vec{c} - \left( \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} ight) \vec{b}$.

The 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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Let $\vec{a}=-\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-3\hat{k}$, $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}=\vec{c}\times\vec{a}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :

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