# Thread: Finding The Equation To A Plane

1. ## Finding The Equation To A Plane

Hello,

The problem I am working on is example number two, given on this webpage
http://www2.gcc.edu/dept/math/facult...e_examples.pdf

What I don't quite understand is, how did they find the point $\displaystyle (-1,0, \sqrt{3})$, without resorting to a guess-and-check method?

2. ## Re: Finding The Equation To A Plane

The article says that "If the plane makes an angle of $\displaystyle \frac{\pi}{6}$ with the x-axis, then the normal vector makes an angle of $\displaystyle \frac{\pi}{6}+ \frac{\pi}{2}= \frac{2\pi}{3}$" (adding the $\displaystyle \pi/2$ for the right angle of the normal vector). The x-coordinate of such a vector is $\displaystyle cos(2\pi/3)= -1/2$, the y-coordinate is 0 because the plane contains the y-axis, and the z coordinate is $\displaystyle sin(2\pi/3)= \sqrt{3}/2)$. That would give vector $\displaystyle <-1/2, 0, \sqrt{3}/2>$. They have simplified slightly by multiplying by 2 which changes the length but not the direction of the vector.