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(D) We are given $\vec{a} + \vec{b} = \mu \vec{p}$,which implies $\vec{a} = \mu \vec{p} - \vec{b}$.We need to evaluate the expression $E = |(\vec{a} \

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$|\vec{p} \cdot \vec{q}|$

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(D) We are given $\vec{a} + \vec{b} = \mu \vec{p}$,which implies $\vec{a} = \mu \vec{p} - \vec{b}$.We need to evaluate the expression $E = |(\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a}|$.Using the vector triple product identity $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w}$,we can rewrite the expression as:$E = |(\vec{p} 	imes \vec{a}) 	imes \vec{q}|$.Substituting $\vec{a} = \mu \vec{p} - \vec{b}$ into the expression:$E = |(\vec{p} 	imes (\mu \vec{p} - \vec{b})) 	imes \vec{q}| = |(\mu (\vec{p} 	imes \vec{p}) - (\vec{p} 	imes \vec{b})) 	imes \vec{q}|$.Since $\vec{p} 	imes \vec{p} = 0$,this simplifies to:$E = |-(\vec{p} 	imes \vec{b}) 	imes \vec{q}| = |(\vec{b} 	imes \vec{p}) 	imes \vec{q}|$.Using the vector triple product identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$:$E = |(\vec{b} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{b}|$.Given $\vec{b} \cdot \vec{q} = 0$ and $|\vec{b}| = 1$,we get:$E = |0 \cdot \vec{p} - (\vec{p} \cdot \vec{q}) \vec{b}| = |-(\vec{p} \cdot \vec{q}) \vec{b}| = |\vec{p} \cdot \vec{q}| \cdot |\vec{b}| = |\vec{p} \cdot \vec{q}| \cdot 1 = |\vec{p} \cdot \vec{q}|$.

Given that $\vec{a}, \vec{b}, \vec{p}$,and $\vec{q}$ are four vectors such that $\vec{a} + \vec{b} = \mu \vec{p}$,$\vec{b} \cdot \vec{q} = 0$,and $|\vec{b}|^2 = 1$,then $|(\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a}|$ is equal to:

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