Let overline{a}=alpha hat{i}+3 hat{j}-hat{k}, | vedclass

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(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|} = \frac{10}{3}$.$|\overline

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$3$

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(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|} = \frac{10}{3}$.$|\overline{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = 3$.So,$\frac{(\alpha \hat{i} + 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{3} = \frac{10}{3}$.$\alpha + 6 + 2 = 10 \Rightarrow \alpha = 2$.Now,$\overline{b} 	imes \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -\beta & 4 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.Given $\overline{b} 	imes \overline{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$,comparing components:$6 + \beta = 7 \Rightarrow \beta = 1$.Thus,$\alpha^2 + \beta^2 - \alpha\beta = 2^2 + 1^2 - (2)(1) = 4 + 1 - 2 = 3$.

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