Question

Find the area of the parallelogram with vertices: (2,2,0),(5,3,0),(4,7,0),and(7,8,0).

Answer 1

Answer:

13 square units

Step-by-step explanation:

Let  the vertices of parallelogram are

A(2,2,0),B(5,3,0),C(4,7,0) and D(7,8,0)

Side AB=$B-A=<5,3,0>-<2,2,0>=<3,1,0>$

Side AD=D-A=$<7,8,0>-<2,2,0>=<5,6,0>$

We know that

Area of parallelogram=Magnitude of product of adjacent sides

$AB\times AD=\begin{vmatrix}i&j&k\\3&1&0\\5&6&0\end{vmatrix}$

$AB\times AD=<0,0,1>$

$\mid AB\times AD\mid=\sqrt{x^2+y^2+z^2}$

$\mid AB\times AD\mid=\sqrt{0+0+(13)^2}$

Where x=Component along x-axis

y=Component along y- axis

z=Component along z-axis

$\mid AB\times AD\mid=13$Square units

Hence, the area of parallelogram=13 square units