The area of the given parallelogram with vertices (0,0), (3,7), (7,4), and (10,11) is 37 sq. units.
The area of the parallelogram is
A = |[tex]\vec a[/tex] × [tex]\vec b[/tex]| sq. units
Where '[tex]\vec a[/tex]' and '[tex]\vec b[/tex]' are the adjacents vectors of the parallelogram
[tex]\vec a[/tex] × [tex]\vec b[/tex] - is the cross product
|[tex]\vec a[/tex] × [tex]\vec b[/tex]| - is the absolute value of the magnitude of the cross product
It is given that,
A parallelogram with vertices A(0,0), B(3,7), C(10,11), and D(7,4)
Then, the adjacent vectors are calculated by
[tex]\vec a[/tex] = [tex]\vec {AD}[/tex]
= [tex]\vec D-\vec A[/tex]
= <7-0, 4-0>
= <7, 4>
[tex]\vec b[/tex] = [tex]\vec {AB}[/tex]
= [tex]\vec B-\vec A[/tex]
= <3-0, 7-0>
= <3, 7>
So, the cross product of these two vectors is,
[tex]\vec a \times \vec b[/tex] = [tex]\left[\begin{array}{ccc}i&j&k\\3&7&0\\7&4&0\end{array}\right][/tex]
= i(0) - j(0) + k(12 - 49)
= -37k
The magnitude of the cross product = [tex]\sqrt{(-37)^2}[/tex] = 37
Then, the area of the given parallelogram is
A = |[tex]\vec a[/tex] × [tex]\vec b[/tex]| = |37| = 37 sq. units.
Hence, the area of the given parallelogram is 37 sq. units.
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