If vec{a} is perpendicular to vec{b} and vec{ | vedclass

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(A) Given the equation $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$.Since $\vec{a} \perp \vec{b}$,we have $\vec{a} \cdot \vec{b} = 0$.Taking

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$\frac{\vec{c}}{p} - \left( \frac{\vec{b} \cdot \vec{c}}{p^2} \right)\vec{a}$

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(A) Given the equation $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$.Since $\vec{a} \perp \vec{b}$,we have $\vec{a} \cdot \vec{b} = 0$.Taking the dot product of the given equation with $\vec{b}$:$p(\vec{r} \cdot \vec{b}) + (\vec{r} \cdot \vec{b})(\vec{a} \cdot \vec{b}) = \vec{c} \cdot \vec{b}$.Since $\vec{a} \cdot \vec{b} = 0$,this simplifies to $p(\vec{r} \cdot \vec{b}) = \vec{c} \cdot \vec{b}$,which implies $\vec{r} \cdot \vec{b} = \frac{\vec{c} \cdot \vec{b}}{p}$.Now,substitute this back into the original equation:$p\vec{r} + \left( \frac{\vec{c} \cdot \vec{b}}{p} \right)\vec{a} = \vec{c}$.$p\vec{r} = \vec{c} - \left( \frac{\vec{c} \cdot \vec{b}}{p} \right)\vec{a}$.Dividing by $p$,we get $\vec{r} = \frac{\vec{c}}{p} - \left( \frac{\vec{c} \cdot \vec{b}}{p^2} \right)\vec{a}$.

If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{r}$ is a non-zero vector such that $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$,then $\vec{r} = $

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