Finding the direction vector with only direction angles..

nicole

Hey everybody! Thanks for any help!

If I am told a line has direction angles of 60, 45 and 60 and passes through the point (-2, 1, 3). How would I go about figuring out the symmetric equations of the line..

Relatively simple question but I am a tad confused. HELP!
THANKS AGAIN!

HallsofIvy

If [tex]\theta[/tex], [tex]\phi[/tex], and [tex]\psi[/tex] are the "direction angles", then [tex]cos(\theta)[/tex], [tex]cos(\phi)[/tex], and [tex]cos(\psi)[/tex], the "direction cosines", form a unit vector in that direction.
cos(60)= 1/2, cos(45)= [tex]\frac{\sqrt{2}}{2}[/tex] so a unit vector in the direction with direction angles 60, 45, 60 (degrees- it would be good idea to say that explicitely!) is [tex]\frac{1}{2}i+ \frac{\sqrt{2}}{2}j+ \frac{1}{2}k[/tex] and parametric equations for a line in that direction, passing through (-2, 1, 3) would be [tex]x= \frac{1}{2}t- 2[/tex], [tex]y= \frac{\sqrt{2}}{2}t+ 2[/tex], [tex]z= \frac{1}{2}t+ 3[/tex].