The area of a parallelogram whose diagonals a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

The adjacent sides of a parallelogram are $\bar{a} = \hat{j} + 2\hat{k}$ and $\bar{b} = \hat{i} + 2\hat{j}$. Find its area.

$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

Let $\bar{a}=\bar{i}-\bar{j}+\bar{k}, \bar{b}=\bar{i}-2\bar{j}-2\bar{k}, \bar{c}=6\bar{i}+3\bar{j}-2\bar{k}$ be three vectors. If $\bar{d}$ is a vector perpendicular to both $\bar{a}$ and $\bar{b}$, and $|\bar{d} \times \bar{c}|=14$, then $|\bar{d} \cdot \bar{c}|=$

The unit vector perpendicular to the vectors $6i + 2j + 3k$ and $3i - 6j - 2k$ is

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is a unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then the unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module