Let ABCD be a parallelogram where overrightar | vedclass

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Question

(A) Given that $|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and the area of the parallelogram is $|\overrightarrow{a} 	imes \overrightarrow{b}|

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(A) Given that $|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and the area of the parallelogram is $|\overrightarrow{a} 	imes \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin 	heta = 4 \sin 	heta$.The dot product is $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos 	heta = 4 \cos 	heta$.The given equation is $|\overrightarrow{a} 	imes \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$.Substituting the values,we get $4 \sin 	heta + 4 \cos 	heta = \sqrt{2} (2)(2) = 4\sqrt{2}$.Dividing by $4$,we get $\sin 	heta + \cos 	heta = \sqrt{2}$.Multiplying by $\frac{1}{\sqrt{2}}$,we get $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta = 1$,which is $\sin(	heta + \frac{\pi}{4}) = 1$.Thus,$	heta + \frac{\pi}{4} = \frac{\pi}{2}$,which implies $	heta = \frac{\pi}{4}$.The area of the parallelogram is $|\overrightarrow{a} 	imes \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin 	heta = (2)(2) \sin(\frac{\pi}{4}) = 4 	imes \frac{1}{\sqrt{2}} = 2\sqrt{2}$ square units.

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