# Trigonometry - Finding equation of line

1. Jun 13, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Consider P(2,1,3) and the line L: $$\begin{Bmatrix}x-y-5=0 \\z-1=0 \end{matrix}$$
a) Find the equation of the line that pass through P and cuts L on a perpendicular angle.

b)Finds the points A and B in L in a way that PAB being an equilateral triangle.
2. Relevant equations
I've solved A, and I'm trying to solve b. I think that it can't be much difficult, cause I know all of the interior angles of PAB. I have a trigonometric problem. I've calculated the distance from L to P, so I got the high of the triangle. So, I know that $$sin 60º=h$$, I got $$h=\sqrt[ ]{22}$$, so what I know is that $$sin 60º=\sqrt[ ]{22}$$, so the cosine would be the half of the base. How should I get the base?

3. The attempt at a solution
I've found L2: $$\begin{Bmatrix}x=2+5\mu \\y=1+7\mu \\z=3+4\mu \end{matrix}$$. So, I need to find the points A and B on L.

2. Jun 13, 2010

### HallsofIvy

Re: Trigonometry

No, sin 60º is the height divided by the hypotenuse, the distance from P to either A or B, not just h. And since you do not yet know what A or B are, you do not know that distance.
Fortunately, since $$sin(60º)= \sqrt{3}/{2}= h/hypotenuse$$, $$hypotenuse= 2h/\sqrt{3}$$. And, of course, the base is equal to that so the distance from the foot of the perpendicular to the line from P is half the length of the hypotenuse.