I've tried this problem several times, following a help guide but still can't seem to get it right. Help please?
Find the distance between the skew lines with the given parametric equations.
x = 3 + t, y = 2 + 6t, z = 2t
x = 2 + 4s, y = 4 + 13s, z = -1 + 6s.
I got 3.104 when I did it (68/rad20) but that seems to be wrong. What should I do to solve it correctly?
Find the distance between the skew lines with the given parametric equations.
x = 3 + t, y = 2 + 6t, z = 2t
x = 2 + 4s, y = 4 + 13s, z = -1 + 6s.
I got 3.104 when I did it (68/rad20) but that seems to be wrong. What should I do to solve it correctly?
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L1: x = 3+t y=2+6t z=2t
L2: x = 2+4s y=4+13s z=-1+6s
Take skew line 1, and cross its direction vector with the direction vector of the other line to form a plane that is parallel to line 2.
m1 = (1,6,2)
m2 = (4,13,6)
Plane: 10x+2y-11z+k=0
Plug in a point from skew line 1. (3,2,0)
10(3)+2(2)-11(0)+k=0
30+4+k=0
k= -34
Your plane equation is 10x+2y-11z-34=0.
Using the distance formula, find the distance between this plane and a point in line 2. (2,4,-1)
distance = abs[10(2)+2(4)-11(-1)-34]/sqrt(10^2+2^2+…
= abs[20+8+11-34]/sqrt(100+4+121)
=[5]/sqrt(225)
=5/15
=1/3 units or about 0.33 units << Shortest distance
You can switch the lines and do it and get the same answer too.
L2: x = 2+4s y=4+13s z=-1+6s
Take skew line 1, and cross its direction vector with the direction vector of the other line to form a plane that is parallel to line 2.
m1 = (1,6,2)
m2 = (4,13,6)
Plane: 10x+2y-11z+k=0
Plug in a point from skew line 1. (3,2,0)
10(3)+2(2)-11(0)+k=0
30+4+k=0
k= -34
Your plane equation is 10x+2y-11z-34=0.
Using the distance formula, find the distance between this plane and a point in line 2. (2,4,-1)
distance = abs[10(2)+2(4)-11(-1)-34]/sqrt(10^2+2^2+…
= abs[20+8+11-34]/sqrt(100+4+121)
=[5]/sqrt(225)
=5/15
=1/3 units or about 0.33 units << Shortest distance
You can switch the lines and do it and get the same answer too.