Answer:
Verified
Area = 13.12 square units.
Step-by-step explanation:
Let the given points / vertices of the parallelogram be represented as follows:
A(2,-1,1),
B(5, 1,4),
C(0,1,1),
D(3,3,4)
In vector notation, we can have;
A = 2i - j + k
B = 5i + j + 4k
C = 0i + j + k
D = 3i + 3j +4k
One of the ways to prove that a quadrilateral is a parallelogram is to show that both pairs of opposite sides are parallel.
(i) Now, let's find the various sides of the assumed parallelogram. These sides are:
AB = B - A = [5i + j + 4k] - [2i - j + k] open the brackets
AB = 5i + j + 4k - 2i + j - k collect like terms and solve
AB = 5i - 2i + j + j - k + 4k
AB = 3i + 2j+ 3k
BC = C - B = [0i + j + k] - [5i + j + 4k] open the brackets
BC = 0i + j + k - 5i - j - 4k collect like terms and solve
BC = 0i - 5i + j - j + k - 4k
BC = -5i + 0j - 3k
CD = D - C = [3i + 3j +4k] - [0i + j + k] open the brackets
CD = 3i + 3j + 4k - 0i - j - k collect like terms and solve
CD = 3i - 0i + 3j - j + 4k - k
CD = 3i + 2j + 3k
DA = A - D = [2i - j + k] - [3i + 3j +4k] open the brackets
DA = 2i - j + k - 3i - 3j - 4k collect like terms and solve
DA = 2i - 3i - j - 3j + k - 4k
DA = - i - 4j - 3k
AC = C - A = [0i + j + k] - [2i - j + k] open the brackets
AC = 0i + j + k - 2i + j - k collect like terms and solve
AC = 0i - 2i + j + j + k - k
AC = - 2i + 2j +0k
BD = D - B = [3i + 3j + 4k] - [5i + j + 4k] open the brackets
BD = 3i + 3j + 4k - 5i - j - 4k collect like terms and solve
BD = 3i - 5i + 3j - j + 4k - 4k
BD = - 2i + 2j + 0k
(ii) From the results in (i) above, it has been shown that;
AB is equal to CD, and that implies that AB is parallel to CD. i.e
AB = CD => AB || CD
Also,
AC is equal to BD, and that implies that AC is parallel to BD. i.e
AC = BD => AC || BD
(iii) Therefore, ABDC is a parallelogram since its opposite sides are equal and parallel.
(B) Now let's calculate the area of the parallelogram.
To calculate the area, we find the magnitude of the cross product between any two adjacent sides.
In this case, we choose sides AC and AB.
Area = | AC x AB |
Where;
[tex]AC X AB = \left[\begin{array}{ccc}i&j&k\\-2&2&0\\3&2&3\end{array}\right][/tex]
AC X AB = i(6 - 0) - j(-6 - 0) + k(-4 -6)
AC X AB = 6i + 6j - 10k
|AC X AB| = [tex]\sqrt{6^2 + 6^2 + (-10)^2} \\[/tex]
|AC X AB| = [tex]\sqrt{36 + 36 + 100} \\[/tex]
|AC X AB| = [tex]\sqrt{172} \\[/tex]
|AC X AB| = 13.12
Therefore the area is 13.12 square units.
PS: The diagram showing this parallelogram has been attached to this response.
