Let vec{a} = alpha hat{i} + hat{j} - hat{k} a | vedclass

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(D) First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 1 & -

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$7$

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(D) First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 1 & -1 \ 2 & 1 & -\alpha \end{vmatrix} = \hat{i}(-\alpha + 1) - \hat{j}(-\alpha^2 + 2) + \hat{k}(\alpha - 2) = (1 - \alpha) \hat{i} + (\alpha^2 - 2) \hat{j} + (\alpha - 2) \hat{k}$.Let $\vec{v} = \vec{a} 	imes \vec{b}$ and $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$. The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.The projection of $\vec{v}$ on $\vec{c}$ is given by $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = 30$.$\vec{v} \cdot \vec{c} = (1 - \alpha)(-1) + (\alpha^2 - 2)(2) + (\alpha - 2)(-2) = -1 + \alpha + 2\alpha^2 - 4 - 2\alpha + 4 = 2\alpha^2 - \alpha - 1$.So,$\frac{2\alpha^2 - \alpha - 1}{3} = 30 \implies 2\alpha^2 - \alpha - 1 = 90 \implies 2\alpha^2 - \alpha - 91 = 0$.Solving the quadratic equation $2\alpha^2 - 14\alpha + 13\alpha - 91 = 0 \implies 2\alpha(\alpha - 7) + 13(\alpha - 7) = 0$.Thus,$(\alpha - 7)(2\alpha + 13) = 0$,which gives $\alpha = 7$ or $\alpha = -\frac{13}{2}$.Since $\alpha > 0$,we have $\alpha = 7$.

Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \alpha \hat{k}$,where $\alpha > 0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$ is $30$,then $\alpha$ is equal to:

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