First slide

Straight lines in 3D

Question

The vertices of a triangle are $A (1, 0, 2), B (2, 3, 0), C (0, 0, 4)$ then the symmetric form of the line passing through centroid of the triangle and one vertex $B$ is

Easy

A

$\frac{x - 2}{- 1} = \frac{y - 3}{1} = \frac{z}{2}$
B

$\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{- 2}$
C

$\frac{x - 1}{2} = \frac{y - 1}{1} = \frac{z - 1}{- 2}$
D

$\frac{x}{2} = \frac{y}{2} = \frac{z - 4}{1}$

Solution

The centroid of the triangle formed by the points $A (1, 0, 2), B (2, 3, 0), C (0, 0, 4)$ is $G = (\frac{1 + 2}{3}, \frac{3}{3}, \frac{2 + 4}{3}) = (1, 1, 2)$
The equation of the line joining two points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ is $\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - y_{1}}$
The equation of the line joining two points $(1, 1, 2)$ and $(2, 3, 0)$ is $\frac{x - 1}{2 - 1} = \frac{y - 1}{3 - 1} = \frac{z - 2}{0 - 2}$
This can be rewrite as $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{- 2}$

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