Answer to Question #310469 in Geometry for Alex

Question

The following points define quadrilateral ABCD:

A (1, -5, 2)

B (8, -5, 7)

C (-13, 10, 7)

D (8, -2, -2)

Prove wether this shape is best described as a:

a) Square

b) Rectangle

c) Rhombus

d) Parallelogram

e) Kite

f) None of the above

Answer 1

Solution

The four points are

A (1, -5, 2)

B (8, -5, 7)

C (-13, 10, 7)

D (8, -2, -2)

The distance between the points A and B is

$|AB|=\sqrt{(8-1)^2+(-5+5)^2+(7-2)^2}$

$|AB|=\sqrt{74}$

The distance between the points A and C is

$|AC|=\sqrt{(-13-1)^2+(10+5)^2+(7-2)^2}$

$|AC|=\sqrt{446}$

Similarly

$|AD|=\sqrt{(8-1)^2+(-2+5)^2+(-2-2)^2}=\sqrt{74}$

$|BC|=\sqrt{(-13-8)^2+(10+5)^2+(7-7)^2}=\sqrt{666}$

$|BD|=\sqrt{(8-8)^2+(-2+5)^2+(-2-7)^2}=\sqrt{90}$

$|CD|=\sqrt{(8+13)^2+(-2-10)^2+(-2-7)^2}=\sqrt{666}$

From above we can see that

$|BC|=|CD|=\sqrt{666}$ and $|AB|=|AD|=\sqrt{74}$

It seems its a kite, but the four points are not coplanar, so it can't be a kite because a kite is a plane. It can't be a square or rhombus because no four sides are equal.

It can't be a rectangle or parallelogram, because opposite sides are not equal.

Hence the correct choice is (f) None of the above

The four points have been plotted and they show that they are not coplanar.