First slide

Introduction to 3-D Geometry

Question

The $x, y, z$ coordinates of each vertex of a triangle are in geometric progression .The $x$ and $y$ coordinates of the centroid of the triangle are 1 and 2 respectively. The distance of the centroid from the origin is.

Easy

A

$\sqrt{7}$
B

$\sqrt{21}$
C

$\sqrt{35}$
D

$\sqrt{32}$

Solution

Given $x, y, z$ coordinates of each vertex of a triangle are in geometric progression
and the centroid of the triangle is $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3}) = (1, 2, k)$
it implies that $x_{1} + x_{2} + x_{3} = 3, y_{1} + y_{2} + y_{3} = 6$
suppose that $r$ is common ratio of geometric progression, $r (x_{1} + x_{2} + x_{3}) = 6 \Rightarrow r (3) = 6 \Rightarrow r = 2$
Hence, $z_{1} + z_{2} + z_{3} = r^{2} (x_{1} + x_{2} + x_{3}) = 4 (3) = 12$
The centroid of the triangle is $(1, 2, 4)$
The distance of the centroid from origin is $\sqrt{1 + 4 + 16} = \sqrt{21}$

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