Question

Verify that the points are the vertices of a parallelogram and find its area. (2,-1,1), (5, 1,4), (0,1,1), (3,3,4)

Answer 1

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;

$AC X AB = \left[\begin{array}{ccc}i&j&k\\-2&2&0\\3&2&3\end{array}\right]$

AC X AB = i(6 - 0) - j(-6 - 0) + k(-4 -6)

AC X AB = 6i + 6j - 10k

|AC X AB| = $\sqrt{6^2 + 6^2 + (-10)^2}$

|AC X AB| = $\sqrt{36 + 36 + 100}$

|AC X AB| = $\sqrt{172}$

|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.