Let overrightarrow{a}=hat{i}+2hat{j}+hat{k} a | vedclass

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(B) We need to evaluate the expression $\vec{a} \cdot ((\vec{c} 	imes \vec{b})-\vec{b}-\vec{c})$.Using the distributive property of the dot product,w

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$24$

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(B) We need to evaluate the expression $\vec{a} \cdot ((\vec{c} 	imes \vec{b})-\vec{b}-\vec{c})$.Using the distributive property of the dot product,we get:$\vec{a} \cdot ((\vec{c} 	imes \vec{b})-\vec{b}-\vec{c}) = \vec{a} \cdot (\vec{c} 	imes \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} \quad ........(i)$Given $\vec{a} 	imes \vec{c} = \vec{b}$.Taking the dot product with $\vec{b}$ on both sides:$(\vec{a} 	imes \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b} = |\vec{b}|^2$.Since $\vec{b} = 3\hat{i}-3\hat{j}+3\hat{k}$,we have $|\vec{b}|^2 = 3^2 + (-3)^2 + 3^2 = 9+9+9 = 27$.Thus,$\vec{a} \cdot (\vec{c} 	imes \vec{b}) = [\vec{a} \vec{c} \vec{b}] = (\vec{a} 	imes \vec{c}) \cdot \vec{b} = 27 \quad ........(ii)$Now,calculate $\vec{a} \cdot \vec{b}$:$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0 \quad ........(iii)$Given $\vec{a} \cdot \vec{c} = 3 \quad ........(iv)$Substituting values from $(ii), (iii),$ and $(iv)$ into $(i)$:$27 - 0 - 3 = 24$.

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