Find the distance between the skew lines
(x-1)/2 = (y + 2)/3= (z- 3)/5 and x/3 = (y + 1)/4 = z/2
3 answers
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(x-1)/2 = (y + 2)/3= (z- 3)/5
x = 1 + 2t
y = -2 + 3t
z = 3 + 5t
x/3 = (y + 1)/4 = z/2
x = 0 + 3s
y = -1 + 4s
z = 0 + 2s
we need a normal to <2,3,5> and <3,4,2>
which would be <14,-11,1> , (I assume you know how to find the cross-product)
Let P(1+2t, -2+3t,3+5t) be the point on the first line
let Q(3s, -1+4s,2s) be the point on the second line
then vector PQ = (1+2t-3s, -2+3t + 1-4s, 3+5t-2s) = (2t-3s+1, 3t-4s-1, 5t-2s+3)
(2t-3s+1, 3t-4s-1, 5t-2s+3)dot(2,3,5) = 0
4t-6s+2 + 9t-12s-3 + 25t-10s+15 = 0
38t -28s = -14 #1
(2t-3s+1, 3t-4s-1, 5t-2s+3)dot(3,4,2) = 0
6t-9s+3 + 12t-16s-4 + 10t -4s+6 = 0
28t - 29s = -5 #2
Now solve these two equations for t and s
which let's you find the vector PQ
the magnitude of PQ would be your shortest distance.
Unless I made an arithmetic error, this looks pretty messy
however, I am confident of my method
x = 1 + 2t
y = -2 + 3t
z = 3 + 5t
x/3 = (y + 1)/4 = z/2
x = 0 + 3s
y = -1 + 4s
z = 0 + 2s
we need a normal to <2,3,5> and <3,4,2>
which would be <14,-11,1> , (I assume you know how to find the cross-product)
Let P(1+2t, -2+3t,3+5t) be the point on the first line
let Q(3s, -1+4s,2s) be the point on the second line
then vector PQ = (1+2t-3s, -2+3t + 1-4s, 3+5t-2s) = (2t-3s+1, 3t-4s-1, 5t-2s+3)
(2t-3s+1, 3t-4s-1, 5t-2s+3)dot(2,3,5) = 0
4t-6s+2 + 9t-12s-3 + 25t-10s+15 = 0
38t -28s = -14 #1
(2t-3s+1, 3t-4s-1, 5t-2s+3)dot(3,4,2) = 0
6t-9s+3 + 12t-16s-4 + 10t -4s+6 = 0
28t - 29s = -5 #2
Now solve these two equations for t and s
which let's you find the vector PQ
the magnitude of PQ would be your shortest distance.
Unless I made an arithmetic error, this looks pretty messy
however, I am confident of my method
Just thought of a very simple method
I already found the normal to both lines in my other solution as < 14, -11, 1 >
So a plane perpendicular to the first line is 14x - 11y + z = c
but (1, -2, 3) lies on this plane, so
14 +22 + 3 = c = 39
equation of plane containing the first line is 14x - 11y + z - 39 = 0
A plane containing the 2nd line must be parallel to the first plane, and any point
on that 2nd plane must be equidistant from the first plane
But we know ( 0, -1, 0 ) lies on that 2nd plane
distance between the two planes, or the distance between the lines:
| 14(0) - 11(-1) + 0 - 39|/√(14^2 + 11^2 + 1^2)
= 28/√318
I already found the normal to both lines in my other solution as < 14, -11, 1 >
So a plane perpendicular to the first line is 14x - 11y + z = c
but (1, -2, 3) lies on this plane, so
14 +22 + 3 = c = 39
equation of plane containing the first line is 14x - 11y + z - 39 = 0
A plane containing the 2nd line must be parallel to the first plane, and any point
on that 2nd plane must be equidistant from the first plane
But we know ( 0, -1, 0 ) lies on that 2nd plane
distance between the two planes, or the distance between the lines:
| 14(0) - 11(-1) + 0 - 39|/√(14^2 + 11^2 + 1^2)
= 28/√318