..? can anyone help?
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!
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)