First slide

Introduction to 3-D Geometry

Question

The points $(- 1, 0, 1), (0, 1, - 1), (1, - 1, 0), (\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}})$ form a

Moderate

A

Parallelogram
B

Square
C

Tetrahedron with equilateral triangle as base
D

rectangle

Solution

Given points are $A (- 1, 0, 1), B (0, 1, - 1), C (1, - 1, 0), D (\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}})$
The lengths of the sides
$\begin{matrix} A B = \sqrt{1 + 1 + 4} = \sqrt{6} \\ B C = \sqrt{1 + 4 + 1} = \sqrt{6} \\ A C = \sqrt{4 + 1 + 1} = \sqrt{6} \end{matrix}$
The vertices A, B, C forms an equilateral triangle, and the fourth vertex D does not lies in the plane determined by A, B, C
Hence, the figure formed by these four vertices is tetrahedron.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App