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