# Vectors

• Aug 29th 2009, 05:20 PM
skeske1234
Vectors
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
(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!
• Aug 30th 2009, 06:20 AM
skeske1234
..? can anyone help?
• Aug 30th 2009, 06:40 AM
pedrosorio
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:

(v1,v2,v3) . (1,2,-2) = 0 <=> v1+2v2-2v3 = 0 (1)

They must intersect, therefore:

(-5,-4,2) + k(v1,v2,v3) = (7,-13,8) + t(1,2,-2) <=>

-5 + kv1 = 7 + t
-4 + kv2 = -13 + 2t
2 + kv3 = 8 - 2t

v1 = (12 + t) / k
v2 = (-9 + 2t) / k
v3 = (6 - 2t) / k

Taking (1) we have, 12 + t + 2(-9 + 2t) - 2(6 - 2t) = 0 <=>

-18 + 9t = 0 <=> t = 2

picking k=1,

v1 = 14
v2 = -5
v3 = 2

Therefore the line has equation:

(-5,-4,2) + k (14,-5,2)

And the point of intersection is given when k=1 (and t=2):

(9,-9,4)
• Aug 30th 2009, 08:50 AM
skeske1234
so should POI be (9,-9,4) and not (-9,-9,4)?
• Aug 30th 2009, 08:52 AM
pedrosorio
Yeah, sure, my mistake.