Answer:
Area = 13.15 square units
Step-by-step explanation:
Let the given vertices be represented as follows:
A(2, -1, 1) = 2i - j + k
B(5, 1, 4) = 5i + j + 4k
C(0, 1, 1) = 0i + j + k
D(3, 3, 4) = 3i + 3j + 4k
(i) Let's calculate the vectors of all the sides:
AB = B - A = (5i + j + 4k) - (2i - j + k)
AB = 5i + j + 4k - 2i + j - k [Collect like terms]
AB = 3i + 2j + 3k
BC = C - B = (0i + j + k) - (5i + j + 4k)
BC = 0i + j + k - 5i - j - 4k [Collect like terms]
BC = -5i + 0j - 3k
CD = D - C = (3i + 3j + 4k) - (0i + j + k)
CD = 3i + 3j + 4k - 0i - j - k [Collect like terms]
CD = 3i + 2j + 3k
DA = A - D = (2i - j + k) - (3i + 3j + 4k)
DA = 2i - j + k - 3i - 3j - 4k [Collect like terms]
DA = -i - 4j - 3k
AC = C - A = (0i + j + k) - (2i - j + k)
AC = 0i + j + k - 2i + j - k [Collect like terms]
AC = -2i + 2j
BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)
BD = 3i + 3j + 4k - 5i - j - 4k [Collect like terms]
BD = -2i + 2j
(ii) From the results in (i) above, we can deduce that;
AB = CD This implies that AB || CD [AB is parallel to CD]
AC = BD This implies that AC || BD [AC is parallel to BD]
(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram
<u>Now let's calculate the area</u>
To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.
In this case, we'll choose AB and AC
Area = |AB X AC|
Where;
![AB X AC = \left[\begin{array}{ccc}i&j&k\\3&2&3\\-2&2&0\end{array}\right]](https://tex.z-dn.net/?f=AB%20X%20AC%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C3%262%263%5C%5C-2%262%260%5Cend%7Barray%7D%5Cright%5D)
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AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)
AB X AC = - 6i - 6j + 10k
|AB X AC| = 
|AB X AC| = 
|AB X AC| = 13.15
Area = 13.15 square units.
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<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.
