The area of a parallelogram whose adjacent si | vedclass

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Question

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

Vedclass · Accepted Answer

$2\sqrt{6}$

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,$|\vec{a} 	imes \vec{b}|$. First,we calculate the cross product $\vec{a} 	imes \vec{b}$: $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 0 & -1 & -2 \end{vmatrix}$ $= \hat{i}(3(-2) - 4(-1)) - \hat{j}(2(-2) - 4(0)) + \hat{k}(2(-1) - 3(0))$ $= \hat{i}(-6 + 4) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0)$ $= -2\hat{i} + 4\hat{j} - 2\hat{k}$. Now,we find the magnitude of this vector: $|\vec{a} 	imes \vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2}$ $= \sqrt{4 + 16 + 4} = \sqrt{24} = \sqrt{4 	imes 6} = 2\sqrt{6}$. Thus,the area is $2\sqrt{6}$ sq. units.

The area of a parallelogram whose adjacent sides are $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = -\hat{j} - 2\hat{k}$ is . . . . . . sq. units.

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