Let vec{a}, vec{b}, vec{c}, vec{d} be four ve | vedclass

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(B) Given four vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$.Since $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{c} = 0$.Given that $\vec{b}$ is

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

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(B) Given four vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$.Since $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{c} = 0$.Given that $\vec{b}$ is parallel to $(\vec{c} - \vec{d})$, there exists a scalar $\lambda$ such that $\lambda \vec{b} = \vec{c} - \vec{d}$.Taking the dot product of both sides with $\vec{a}$, we get $\lambda (\vec{a} \cdot \vec{b}) = \vec{a} \cdot (\vec{c} - \vec{d})$.Since $\vec{a} \cdot \vec{c} = 0$, this simplifies to $\lambda (\vec{a} \cdot \vec{b}) = -(\vec{a} \cdot \vec{d})$.Thus, $\lambda = -\frac{\vec{a} \cdot \vec{d}}{\vec{a} \cdot \vec{b}}$.Substituting $\lambda$ back into the equation $\vec{c} = \vec{d} + \lambda \vec{b}$, we obtain $\vec{c} = \vec{d} - \left(\frac{\vec{a} \cdot \vec{d}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$.

Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be four vectors such that $\vec{a}$ is perpendicular only to $\vec{c}$. If the vector $\vec{b}$ is parallel to $(\vec{c}-\vec{d})$, then $\vec{c}$ is equal to:

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