First slide

Introduction to 3-D Geometry

Question

The points A(5, - 1, 1) B(7, - 4,7),C(1, - 6, 10) and D(1 - 3,4) are vertices of a

Easy

A

square
B

rhombus
C

rectangle
D

None of these

Solution

Given that, A(5, - 1, 1) B(7, - 4,7),C(1, - 6, 10) and D(1 - 3,4)
now,
$\begin{array}{l} AB = \sqrt{(7 - 5)^{2} + (- 4 + 1)^{2} + (7 - 1)^{2}} \\ = \sqrt{4 + 9 + 36} = 7 \\ BC = \sqrt{(1 - 7)^{2} + (- 6 + 4)^{2} + (10 - 7)^{2}} \\ = \sqrt{36 + 4 + 9} = 7 \\ CD = \sqrt{(- 1 - 1)^{2} + (- 3 + 6)^{2} + (4 - 10)^{2}} \\ = \sqrt{4 + 9 + 36} = 7 \\ DA = \sqrt{(5 + 1)^{2} + (- 1 + 3)^{2} + (1 - 4)^{2}} \\ = \sqrt{36 + 4 + 9} = 7 \\ ∴ AB = BC = CD = DA = 7, \\ Also, AC = \sqrt{(1 - 5)^{2} + (- 6 + 1)^{2} + (10 - 1)^{2}} \\ = \sqrt{16 + 25 + 81} = \sqrt{122} \\ BD = \sqrt{(- 1 - 7)^{2} + (- 4 + 3)^{2} + (7 - 4)^{2}} \\ = \sqrt{64 + 1 + 9} = \sqrt{74} \end{array}$
Diagonal $AC \neq BD$
Hence, ABCD is not a square, so it is a rhombus.

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