The area of the parallelogram whose diagonals | vedclass

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(B) Let the diagonals of the parallelogram be $\vec{d_1} = 2\vec{a} - \vec{b}$ and $\vec{d_2} = 4\vec{a} - 5\vec{b}.$The area of a parallelogram with

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$\frac{3}{\sqrt{2}}$

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(B) Let the diagonals of the parallelogram be $\vec{d_1} = 2\vec{a} - \vec{b}$ and $\vec{d_2} = 4\vec{a} - 5\vec{b}.$The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|.$First,calculate the cross product: $\vec{d_1} 	imes \vec{d_2} = (2\vec{a} - \vec{b}) 	imes (4\vec{a} - 5\vec{b}).$$= 2\vec{a} 	imes 4\vec{a} - 2\vec{a} 	imes 5\vec{b} - \vec{b} 	imes 4\vec{a} + \vec{b} 	imes 5\vec{b}.$Since $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{b} = 0,$ we have:$= -10(\vec{a} 	imes \vec{b}) - 4(\vec{b} 	imes \vec{a}) = -10(\vec{a} 	imes \vec{b}) + 4(\vec{a} 	imes \vec{b}) = -6(\vec{a} 	imes \vec{b}).$Given that $\vec{a}$ and $\vec{b}$ are unit vectors $(|\vec{a}| = 1, |\vec{b}| = 1)$ and the angle between them is $45^{\circ},$$|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| \sin(45^{\circ}) = 1 	imes 1 	imes \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}.$Thus,$|\vec{d_1} 	imes \vec{d_2}| = |-6(\vec{a} 	imes \vec{b})| = 6 	imes \frac{1}{\sqrt{2}} = 3\sqrt{2}.$Therefore,the area is $\frac{1}{2} 	imes 3\sqrt{2} = \frac{3}{\sqrt{2}}.$

The area of the parallelogram whose diagonals are the vectors $2\vec{a} - \vec{b}$ and $4\vec{a} - 5\vec{b},$ where $\vec{a}$ and $\vec{b}$ are unit vectors forming an angle of $45^{\circ},$ is

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