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(A) 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

$15\sqrt{2}$

Vedclass · Answer

(A) 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} \ 1 & -1 & 3 \ 2 & -7 & 1 \end{vmatrix}$$= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))$$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$$= 20\hat{i} + 5\hat{j} - 5\hat{k}$Now,find the magnitude of this vector:$|\vec{a} 	imes \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2}$$= \sqrt{400 + 25 + 25} = \sqrt{450}$$= \sqrt{225 	imes 2} = 15\sqrt{2}$Thus,the area of the parallelogram is $15\sqrt{2}$ square units.

Two adjacent sides of a parallelogram are given by vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find the area of the parallelogram in square units.

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