show that the quadrilateral with vertices at the following points is a parallelogram and find its area. a = (5,2,0), b = (2,6,1), c = (2,4,7), d = (5,0,6)
Just noticed that Steve answered the same question when you reposted it just a few entries above this
Please avoid reposting the same question, especially when it still shows near the top of the list of questions like it did here.
It avoids unnecessary duplication of solutions.
To show that a quadrilateral is a parallelogram, we need to prove that opposite sides are parallel.
Step 1: Find the vectors that represent the sides of the quadrilateral.
Let's find the vectors AB, BC, CD, and DA.
AB = b - a = (2, 6, 1) - (5, 2, 0) = (-3, 4, 1)
BC = c - b = (2, 4, 7) - (2, 6, 1) = (0, -2, 6)
CD = d - c = (5, 0, 6) - (2, 4, 7) = (3, -4, -1)
DA = a - d = (5, 2, 0) - (5, 0, 6) = (0, 2, -6)
Step 2: Check if opposite sides are parallel.
To do that, we need to check if the cross product of AB and CD is zero and if the cross product of BC and DA is zero.
AB x CD = (-3, 4, 1) x (3, -4, -1)
Using the cross product formula, AB x CD = (4 - (-4), 1 - 3, (-3) - 12) = (8, -2, -15)
Since the cross product is not zero, AB and CD are not parallel.
BC x DA = (0, -2, 6) x (0, 2, -6)
Using the cross product formula, BC x DA = ((-2)(-6) - (2)(6), (0)(-6) - (0)(-2), (0)(2) - (-2)(0)) = (12, 0, 0)
Since the cross product is zero, BC and DA are parallel.
Step 3: If opposite sides are parallel, then the quadrilateral is a parallelogram. Therefore, the given quadrilateral is a parallelogram.
Step 4: Find the area of the parallelogram.
To find the area of a parallelogram, we need to calculate the magnitude of the cross product of two adjacent sides and multiply it by the length of one of the sides.
In this case, we can use AB and BC as adjacent sides.
Area = magnitude(AB x BC) * ||AB||
Area = magnitude((-3, 4, 1) x (0, -2, 6)) * ||(-3, 4, 1)||
Using the cross product formula, |AB x BC| = sqrt((-2)^2 + (3)^2 + (4)^2) = sqrt(29)
||AB|| = sqrt((-3)^2 + (4)^2 + (1)^2) = sqrt(26)
Area = sqrt(29) * sqrt(26) = sqrt(754)
Therefore, the area of the given parallelogram is sqrt(754).
To prove that the given quadrilateral is a parallelogram, we need to show that both pairs of opposite sides are parallel.
1. Calculate the vectors of the two diagonals of the quadrilateral:
- Diagonal AC: vector AC = c - a = (2, 4, 7) - (5, 2, 0) = (-3, 2, 7)
- Diagonal BD: vector BD = d - b = (5, 0, 6) - (2, 6, 1) = (3, -6, 5)
2. Check if the diagonals are parallel:
Two vectors are parallel if their cross product is a zero vector.
Cross product: AC × BD = (-3, 2, 7) × (3, -6, 5)
Cross product: AC × BD = (2*5 - 7*(-6), 7*3 - (-3)*5, (-3)*(-6) - 2*5)
Cross product: AC × BD = (44, 36, -33)
Since the cross product of the diagonals is not the zero vector, the quadrilateral is not a parallelogram. Therefore, we can't find its area.
Please note that if you made an error in providing the coordinates or there was a typo, the calculations may differ, but the process remains the same.
vector ab = (-3,4,1)
vector dc = (3, -4, -1) = vector cd
since their magnitude is the same, and they simply have opposite direction they must be equal and parallel
So the figure is a parallogram
vector cb = (0,-2,6)
area = cross product of ab and cb
= (-3,4,1) x (0,-2,6)
= (26,18,6)
whose magnitude is √(26^2+18^2+6^) = √1036