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

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(B) The position vectors of vertices $A, B, C,$ and $D$ of rectangle $ABCD$ are given as:$\overrightarrow{OA} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{

Vedclass · Accepted Answer

$2$

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(B) The position vectors of vertices $A, B, C,$ and $D$ of rectangle $ABCD$ are given as:$\overrightarrow{OA} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}, \overrightarrow{OB} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$$\overrightarrow{OC} = \hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}, \overrightarrow{OD} = -\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$The adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are calculated as:$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1 - (-1))\hat{i} + (\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} = 2\hat{i}$$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (1 - 1)\hat{i} + (-\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} = -\hat{j}$The area of the rectangle is given by the magnitude of the cross product of its adjacent sides:$	ext{Area} = |\overrightarrow{AB} 	imes \overrightarrow{BC}|$$\overrightarrow{AB} 	imes \overrightarrow{BC} = (2\hat{i}) 	imes (-\hat{j}) = -2(\hat{i} 	imes \hat{j}) = -2\hat{k}$$|\overrightarrow{AB} 	imes \overrightarrow{BC}| = |-2\hat{k}| = 2$Thus,the area of the rectangle is $2$ 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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