The area of the parallelogram formed by vecto | vedclass

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(B) The area of a parallelogram formed by two vectors $\vec{P}$ and $\vec{Q}$ is given by the magnitude of their cross product, i.e.,$	ext{Area} = |\

Vedclass · Accepted Answer

$\sqrt{86}$

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(B) The area of a parallelogram formed by two vectors $\vec{P}$ and $\vec{Q}$ is given by the magnitude of their cross product, i.e.,$	ext{Area} = |\vec{P} 	imes \vec{Q}|$.First, calculate the cross product $\vec{P} 	imes \vec{Q}$ using the determinant method:$\vec{P} 	imes \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 5 \ 3 & -2 & 4 \end{vmatrix}$$= \hat{i}((-1)(4) - (5)(-2)) - \hat{j}((2)(4) - (5)(3)) + \hat{k}((2)(-2) - (-1)(3))$$= \hat{i}(-4 + 10) - \hat{j}(8 - 15) + \hat{k}(-4 + 3)$$= 6\hat{i} + 7\hat{j} - 1\hat{k}$Now, find the magnitude of the resulting vector:$|\vec{P} 	imes \vec{Q}| = \sqrt{(6)^2 + (7)^2 + (-1)^2}$$= \sqrt{36 + 49 + 1}$$= \sqrt{86}$Thus, the area of the parallelogram is $\sqrt{86}$ square units.

The area of the parallelogram formed by vectors $\vec{P} = 2\hat{i} - \hat{j} + 5\hat{k}$ and $\vec{Q} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ is

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