The area of the parallelogram whose sides are | vedclass

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(B) Let the two vectors representing the sides of the parallelogram be $\vec{A} = 0\hat{i} + 1\hat{j} + 3\hat{k}$ and $\vec{B} = 1\hat{i} + 2\hat{j} -

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$\sqrt{59}$ sq. unit

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(B) Let the two vectors representing the sides of the parallelogram be $\vec{A} = 0\hat{i} + 1\hat{j} + 3\hat{k}$ and $\vec{B} = 1\hat{i} + 2\hat{j} - 1\hat{k}$.The area of a parallelogram formed by two vectors $\vec{A}$ and $\vec{B}$ is given by the magnitude of their cross product: $	ext{Area} = |\vec{A} 	imes \vec{B}|$.First,calculate the cross product $\vec{A} 	imes \vec{B}$ using the determinant method:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 3 \ 1 & 2 & -1 \end{vmatrix}$$= \hat{i}((1)(-1) - (3)(2)) - \hat{j}((0)(-1) - (3)(1)) + \hat{k}((0)(2) - (1)(1))$$= \hat{i}(-1 - 6) - \hat{j}(0 - 3) + \hat{k}(0 - 1)$$= -7\hat{i} + 3\hat{j} - 1\hat{k}$.Now,calculate the magnitude of the resulting vector:$|\vec{A} 	imes \vec{B}| = \sqrt{(-7)^2 + (3)^2 + (-1)^2}$$= \sqrt{49 + 9 + 1} = \sqrt{59}$ sq. unit.

The area of the parallelogram whose sides are represented by the vectors $\hat j + 3\hat k$ and $\hat i + 2\hat j - \hat k$ is

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