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(B) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d

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$\frac{3}{\sqrt{2}}$ sq. units

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(B) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$. Given $\vec{d_1} = 2\bar{a} - \bar{b}$ and $\vec{d_2} = 4\bar{a} - 5\bar{b}$. Calculating the cross product: $\vec{d_1} 	imes \vec{d_2} = (2\bar{a} - \bar{b}) 	imes (4\bar{a} - 5\bar{b})$ $= 2\bar{a} 	imes 4\bar{a} - 2\bar{a} 	imes 5\bar{b} - \bar{b} 	imes 4\bar{a} + \bar{b} 	imes 5\bar{b}$ Since $\bar{a} 	imes \bar{a} = 0$ and $\bar{b} 	imes \bar{b} = 0$,this simplifies to: $= -10(\bar{a} 	imes \bar{b}) - 4(\bar{b} 	imes \bar{a})$ Since $\bar{b} 	imes \bar{a} = -(\bar{a} 	imes \bar{b})$,we have: $= -10(\bar{a} 	imes \bar{b}) + 4(\bar{a} 	imes \bar{b}) = -6(\bar{a} 	imes \bar{b})$. The magnitude is $|-6(\bar{a} 	imes \bar{b})| = 6 |\bar{a}| |\bar{b}| \sin(45^{\circ})$. Given $|\bar{a}| = 1$,$|\bar{b}| = 1$,and $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,$|\vec{d_1} 	imes \vec{d_2}| = 6 	imes 1 	imes 1 	imes \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$. Thus,$	ext{Area} = \frac{1}{2} 	imes 3\sqrt{2} = \frac{3}{\sqrt{2}}$ sq. units.

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

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