First slide

Introduction to 3-D Geometry

Question

If $D (2, 1, 0), E (2, 0, 0), F (0, 1, 0)$ are mid points of the sides $B C, C A, A B$ of triangle $A B C$ respectively. Then, the centroid of triangle $A B C$ is

Easy

A

$(\frac{- 1}{3}, \frac{1}{3}, \frac{1}{3})$
B

$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$
C

$(\frac{2}{3}, \frac{1}{3}, \frac{1}{3})$
D

$(\frac{4}{3}, \frac{2}{3}, 0)$

Solution

Given are $D, E, F$ midpoints of the sides of the triangle $A B C$ , so that the centroid of the triangle $A B C$ and the centroid of the triangle $D E F$ are coincident.
Hence, the centroid of the triangle $D E F$ is $(\frac{2 + 2 + 0}{3}, \frac{1 + 0 + 1}{3}, \frac{0 + 0 + 0}{3}) = (\frac{4}{3}, \frac{2}{3}, 0)$
Therefore, the centroid of the triangle $A B C$ is $(\frac{4}{3}, \frac{2}{3}, 0)$

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