Let vec{a}=hat{i}-2hat{j}+hat{k} and vec{b}=h | vedclass

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(C) Given $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{a}$,we can write $\vec{b} 	imes (\vec{c} - \vec{a}) = \vec{0}$.This implies that $\vec{c} - \

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$-\frac{1}{2}$

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(C) Given $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{a}$,we can write $\vec{b} 	imes (\vec{c} - \vec{a}) = \vec{0}$.This implies that $\vec{c} - \vec{a}$ is parallel to $\vec{b}$,so $\vec{c} - \vec{a} = k\vec{b}$ for some scalar $k$.Thus,$\vec{c} = \vec{a} + k\vec{b}$.Given $\vec{c} \cdot \vec{a} = 0$,we substitute $\vec{c}$:$(\vec{a} + k\vec{b}) \cdot \vec{a} = 0 \Rightarrow |\vec{a}|^2 + k(\vec{b} \cdot \vec{a}) = 0$.Calculating magnitudes and dot products:$|\vec{a}|^2 = 1^2 + (-2)^2 + 1^2 = 6$.$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(-2) + (1)(1) = 1 + 2 + 1 = 4$.Substituting these values: $6 + k(4) = 0 \Rightarrow 4k = -6 \Rightarrow k = -\frac{3}{2}$.Now,$\vec{c} \cdot \vec{b} = (\vec{a} + k\vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + k|\vec{b}|^2$.$|\vec{b}|^2 = 1^2 + (-1)^2 + 1^2 = 3$.$\vec{c} \cdot \vec{b} = 4 + (-\frac{3}{2})(3) = 4 - \frac{9}{2} = \frac{8-9}{2} = -\frac{1}{2}$.

Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0,$ then $\vec{c} \cdot \vec{b}$ is equal to

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