Equation of a line

1. Oct 14, 2003

PiRsq

I tried to do this but I dont get the answer as in the book...

A line passes through point (0,4). Its first direction angle is 60°, meaning alpha is 60°.

I found that the second direction angles are 30° and 150° thats is beta is 30° and 150°. But what are the parametric equations of the line for each set of direction angles?

Thx any help appreciated

2. Oct 15, 2003

HallsofIvy

Staff Emeritus
My understanding of "direction angles" for a line is that they are the angles the line makes with each of the coordinate axes. In a two dimensional problem, there are only two direction angles. In this case, since one angle is given as 60 degrees (the angle the line makes with the x-axis) the other angle has to be 30 degrees (in two dimensions, the angles have to add to 90 degrees- direction angles (and direction cosines) are more often used in three or more dimensions). I don't understand why "second direction angles" is plural. I also do not understand what you mean by " the parametric equations of the line for each set of direction angles".

3. Oct 15, 2003

gnome

I agree with HallsofIvy that a third direction angle makes no sense in this problem. So ignoring that ...

A vector along that line would be

So a vector equation for the line in the form
r = r0 + tv (where r0 is the position vector of your point (0,4) and t is the parameter might be
<x,y> = <0,4> + t<1,&radic;(3)> or

and then the parametric equations of the line would be
x = t

Last edited: Oct 15, 2003
4. Oct 15, 2003

gnome

That was probably too involved. Maybe this is a better answer:

Since you know that &alpha; is 60o, you know that the slope of the line is tan60o = &radic;(3)

So, an equation for the line through point (0,4) with slope &radic;(3) is
y - 4 = &radic;(3) * (x - 0)