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(C) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} 	imes \ve

Vedclass · Accepted Answer

$\sqrt{42}$

Vedclass · Answer

(C) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} 	imes \vec{b}|$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 4 \ 1 & -1 & 1 \end{vmatrix}$$= \hat{i}(1(1) - 4(-1)) - \hat{j}(3(1) - 4(1)) + \hat{k}(3(-1) - 1(1))$$= \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1)$$= 5\hat{i} + \hat{j} - 4\hat{k}$Now,calculate the magnitude $|\vec{a} 	imes \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2}$$= \sqrt{25 + 1 + 16} = \sqrt{42}$ sq. units.Thus,the correct option is $C$.

The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.

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