If vec{alpha} = 3hat{i} - hat{k},|vec{beta}| | vedclass

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(D) The area of a parallelogram with adjacent sides $\vec{\alpha}$ and $\vec{\beta}$ is given by $|\vec{\alpha} 	imes \vec{\beta}|$. We know that $|\

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$\sqrt{41}$

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(D) The area of a parallelogram with adjacent sides $\vec{\alpha}$ and $\vec{\beta}$ is given by $|\vec{\alpha} 	imes \vec{\beta}|$. We know that $|\vec{\alpha} 	imes \vec{\beta}|^2 = |\vec{\alpha}|^2 |\vec{\beta}|^2 - (\vec{\alpha} \cdot \vec{\beta})^2$. First,calculate $|\vec{\alpha}|^2$: $|\vec{\alpha}|^2 = 3^2 + 0^2 + (-1)^2 = 9 + 1 = 10$. Given $|\vec{\beta}| = \sqrt{5}$,so $|\vec{\beta}|^2 = 5$. Given $\vec{\alpha} \cdot \vec{\beta} = 3$,so $(\vec{\alpha} \cdot \vec{\beta})^2 = 3^2 = 9$. Now,substitute these values into the formula: $|\vec{\alpha} 	imes \vec{\beta}|^2 = (10)(5) - 9 = 50 - 9 = 41$. Therefore,the area is $|\vec{\alpha} 	imes \vec{\beta}| = \sqrt{41}$.

If $\vec{\alpha} = 3\hat{i} - \hat{k}$,$|\vec{\beta}| = \sqrt{5}$,and $\vec{\alpha} \cdot \vec{\beta} = 3$,then the area of the parallelogram for which $\vec{\alpha}$ and $\vec{\beta}$ are adjacent sides is:

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