Trigonometry - Finding equation of line

[Telemachus]

Homework Statement

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.

Homework 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?

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.

 

[HallsofIvy]

Homework Statement

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.

Homework 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 $$sin 60º=h$$

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.

, 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?

Excuse me, but $$\sqrt{22}$$ is almost 5 and a sine cannot be larger than 1! No, $sin(60º)= \sqrt{3}/2$. You can get that by dividing an equilateral triangle into two right triangles with angles 60º and 30º. If each side of the equilateral triangle has length L, then each right triangle has hypotenuse of length L, one leg, opposite the 30º angle, of length L/2, and the other leg, opposite the 60º angle, of length [tex]\sqrt{3}L/2[/itex], by the Pythagorean theorem.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>The Attempt at a Solution</h2><br /> 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. </div> </div> </blockquote>[/tex]

 

[Telemachus]

Thanks HallsofIvy.

So, what I got is that $$x=\displaystyle\frac{\sqrt[ ]{22}}{sin 60º}$$, and that's the length of the sides of the triangle, right?