Find the equation of the line through the point (-5,-4,2) that intersects the line at vector r=(7,-13,8)+ t(1,2,-2) at 90 degrees. Determine the point of intersection.
So this is my work below, so far:
Point A (-5,-4,2)
Point B (7,-13,8)
therefore vector eqtn is:
(x,y,z)=(-5,-4,2) + t(12,-9,6)
So that is my attempt at part A, now part B:
Point of Intersection
l1: (x,y,z)=(-5,-4,2) + t(12,-9,6)
l1: x=-5+12t, y=-4-9t, z=2+6t
l2: (x-7)/1 = (y+13)/2 = (z-8)/-2
Substitute parametric eqtn into symmetric eqtn
l2: t=1 and t=1
therefore point of intersection when t =1
O.K. Please let me know if I have my answers correct and method. If not, please let me know the correct one and thoroughly explain accordingly. Thank you for your help!
August 30th 2009, 06:20 AM
..? can anyone help?
August 30th 2009, 06:40 AM
If the direction vector of your line goes from point A to point B then the intersection of the lines is obviously at point B. You have to find a line that intersects the other one making an angle of 90º.
To do so, the vectors that define directions have to be perpendicular, that is, their inner product must be 0: