Let overline{a}=hat{i}+2 hat{j}-hat{k} and ov | vedclass

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(D) Given,$\overline{b} 	imes \overline{c} = \overline{b} 	imes \overline{a}$.This implies $\overline{b} 	imes (\overline{c} - \overline{a}) = \ove

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$-\frac{1}{2}$

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(D) Given,$\overline{b} 	imes \overline{c} = \overline{b} 	imes \overline{a}$.This implies $\overline{b} 	imes (\overline{c} - \overline{a}) = \overline{0}$.Thus,$\overline{c} - \overline{a} = \lambda \overline{b}$ for some scalar $\lambda$,so $\overline{c} = \overline{a} + \lambda \overline{b}$.Given $\overline{c} \cdot \overline{a} = 0$,we have $(\overline{a} + \lambda \overline{b}) \cdot \overline{a} = 0$.This gives $|\overline{a}|^2 + \lambda (\overline{b} \cdot \overline{a}) = 0$.Calculating the values: $|\overline{a}|^2 = 1^2 + 2^2 + (-1)^2 = 6$ and $\overline{b} \cdot \overline{a} = (1)(1) + (1)(2) + (-1)(-1) = 1 + 2 + 1 = 4$.Substituting these: $6 + 4\lambda = 0 \Rightarrow \lambda = -\frac{6}{4} = -\frac{3}{2}$.Now,$\overline{c} = \overline{a} - \frac{3}{2} \overline{b}$.Then $\overline{c} \cdot \overline{b} = (\overline{a} - \frac{3}{2} \overline{b}) \cdot \overline{b} = \overline{a} \cdot \overline{b} - \frac{3}{2} |\overline{b}|^2$.We have $\overline{a} \cdot \overline{b} = 4$ and $|\overline{b}|^2 = 1^2 + 1^2 + (-1)^2 = 3$.Therefore,$\overline{c} \cdot \overline{b} = 4 - \frac{3}{2}(3) = 4 - \frac{9}{2} = -\frac{1}{2}$.

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

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