Area of a rectangle having vertices A, B, C a | vedclass

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(C) The position vectors of the vertices are given as:$\vec{A} = -\hat{i} + 0.5\hat{j} + 4\hat{k}$$\vec{B} = \hat{i} + 0.5\hat{j} + 4\hat{k}$$\vec{C}

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$2$

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(C) The position vectors of the vertices are given as:$\vec{A} = -\hat{i} + 0.5\hat{j} + 4\hat{k}$$\vec{B} = \hat{i} + 0.5\hat{j} + 4\hat{k}$$\vec{C} = \hat{i} - 0.5\hat{j} + 4\hat{k}$$\vec{D} = -\hat{i} - 0.5\hat{j} + 4\hat{k}$The length of side $AB$ is given by the magnitude of the vector $\vec{AB} = \vec{B} - \vec{A} = (\hat{i} + 0.5\hat{j} + 4\hat{k}) - (-\hat{i} + 0.5\hat{j} + 4\hat{k}) = 2\hat{i}$.$|AB| = |2\hat{i}| = 2$ units.The length of side $BC$ is given by the magnitude of the vector $\vec{BC} = \vec{C} - \vec{B} = (\hat{i} - 0.5\hat{j} + 4\hat{k}) - (\hat{i} + 0.5\hat{j} + 4\hat{k}) = -1\hat{j}$.$|BC| = |-1\hat{j}| = 1$ unit.The area of the rectangle is given by the product of its adjacent sides:$	ext{Area} = |AB| 	imes |BC| = 2 	imes 1 = 2$ square units.

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$,$\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$,$\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$,respectively is . . . . . . .

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