Question

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Answer 1

Answer:

The coordinates of the required point is $(\frac{17}{3}, 0, \frac{23}{3})$

Step-by-step explanation:

Formuila used:

The equation of line joining $(x_1,y_1,z_1)\:and\: (x_2,y_2,z_2)$ is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The equation of line pases through

(5, 1, 6) and (3, 4, 1) is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\\\\\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\\\\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}$

It meets ZX plane.

That is y=0

$\frac{x-5}{-2}=\frac{0-1}{3}=\frac{z-6}{-5}\\\\\frac{x-5}{-2}=\frac{-1}{3}=\frac{z-6}{-5}$

Now,

$x-5=\frac{2}{3}\\\\x=5+\frac{2}{3}\\\\x=\frac{17}{3}$

and

$\frac{-1}{3}=\frac{z-6}{-5}\\\\\frac{5}{3}=z-6\\\\\frac{5}{3}+6=z\\\\z=\frac{23}{3}$

The required point is

$(\frac{17}{3}, 0, \frac{23}{3})$