If diagonals of a parallelogram are (5hat i - | vedclass

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

Question

(A) 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{

Vedclass · Accepted Answer

$\sqrt{171} 	ext{ unit}^2$

Vedclass · Answer

(A) 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 $\vec{d_1} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{d_2} = 3\hat{i} + 2\hat{j} - \hat{k}$.First,calculate the cross product $\vec{d_1} 	imes \vec{d_2}$:$\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & -4 & 3 \ 3 & 2 & -1 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(-5 - 9) + \hat{k}(10 - (-12)) = -2\hat{i} + 14\hat{j} + 22\hat{k}$.Now,find the magnitude of the cross product:$|\vec{d_1} 	imes \vec{d_2}| = \sqrt{(-2)^2 + 14^2 + 22^2} = \sqrt{4 + 196 + 484} = \sqrt{684}$.Simplify the square root: $\sqrt{684} = \sqrt{36 	imes 19} = 6\sqrt{19}$.Finally,the area is $\frac{1}{2} 	imes 6\sqrt{19} = 3\sqrt{19} = \sqrt{9 	imes 19} = \sqrt{171} 	ext{ unit}^2$.

If diagonals of a parallelogram are $(5\hat i - 4\hat j + 3\hat k)$ and $(3\hat i + 2\hat j - \hat k)$,then its area is

Similar Questions

Find the angle between the vectors $A=2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $B=4 \hat{i}+2 \hat{j}-4 \hat{k}$. (in $^{\circ}$)

The vector component of $\vec{a} = 2\hat{i} + 3\hat{j}$ along the direction of vector $\vec{b} = (\hat{i} + \hat{j})$ is:

Two unit vectors $\hat{a}_{1}$ and $\hat{a}_{2}$ are inclined to each other at an angle $\theta$. If $|\hat{a}_{1}-\hat{a}_{2}|=\sqrt{3}$,then the value of $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$ is:

The resultant of $\vec{A} \times 0$ will be equal to

Explain the kinds of multiplication operations for vectors.

Vedclass Test Series

Exam Paper Generator

Online Exam Module