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:

Vector eqtn:

Point A (-5,-4,2)

Point B (7,-13,8)

d=AB=(12, -9,6)

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)

l2: r=(x,y,z)=(7,-13,8)+t(1,2,-2)

Parametric eqtn:

l1: x=-5+12t, y=-4-9t, z=2+6t

Symmetric eqtn:

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

x=-5+12(1)=7

y=-4-9(1)=-13

z=2+6(1)=8

final answer:

(7,-13,8)

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!