Find the distance between the skew lines

(x-1)/2 = (y + 2)/3= (z- 3)/5 and x/3 = (y + 1)/4 = z/2

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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

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

To find the distance between two skew lines, follow these steps:

Step 1: Write the parametric equations for both lines.
For the first line, given (x-1)/2 = (y + 2)/3 = (z- 3)/5, we can write it as:
x = 2t + 1
y = 3t - 2
z = 5t + 3

For the second line, given x/3 = (y + 1)/4 = z/2, we can write it as:
x = 3s
y = 4s - 1
z = 2s

Step 2: Compute the direction vectors of both lines.
The direction vector (d1) of the first line is obtained by coefficients of t in x, y, and z. Therefore, d1 = <2, 3, 5>.

The direction vector (d2) of the second line is obtained by coefficients of s in x, y, and z. Therefore, d2 = <3, 4, 2>.

Step 3: Find the cross product of the direction vectors.
The cross product of the direction vectors gives the vector orthogonal to both directions, which will be used to find the distance between the skew lines.

Cross product: d1 x d2 = (2i + 3j + 5k) x (3i + 4j + 2k)
= (3*5 - 2*4)i + (2*3 - 5*3)j + (2*4 - 3*4)k
= 7i - 9j - 4k

Step 4: Find the magnitude of the cross product.
Magnitude: |d1 x d2| = √(7^2 + (-9)^2 + (-4)^2)
= √(49 + 81 + 16)
= √146

Therefore, the magnitude of the cross product is √146.

Step 5: Compute the distance between the skew lines.
The distance between the skew lines is given by the formula:
Distance = |(P0 - P1) · n| / |n|

Here, (P0 - P1) is the vector from a point on one line to the other line, and n is the unit vector orthogonal to both direction vectors (d1 and d2).

To find (P0 - P1), we can choose an arbitrary point on one line and the closest point on the other line.

Let's choose a point on the first line, P0 = (1, -2, 3).

The closest point on the second line, P1, can be obtained by projecting P0 onto the line.

To find P1, we can use the formula:
P1 = P0 + [(P0 - P2) · d2 / |d2|^2] * d2
where P2 is any point on the second line.

Let's choose a point on the second line, P2 = (0, -1, 0).

Now, substitute the values into the formula:
P1 = (1, -2, 3) + [(1, -2, 3) - (0, -1, 0)] · <3, 4, 2> / |<3, 4, 2>|^2 * <3, 4, 2>
= (1, -2, 3) + (1, -1, 3) · <3, 4, 2> / |<3, 4, 2>|^2 * <3, 4, 2>
= (1, -2, 3) + (3, -4, 6) / (3^2 + 4^2 + 2^2) * <3, 4, 2>
= (1, -2, 3) + (3, -4, 6) / 29 * <3, 4, 2>
= (1, -2, 3) + (9/29, -12/29, 18/29)
= (1 + 9/29, -2 - 12/29, 3 + 18/29)
= (38/29, -58/29, 95/29)

Now, substitute the values into the distance formula:
Distance = |(P0 - P1) · n| / |n|
= |<(1, -2, 3) - (38/29, -58/29, 95/29)> · <7, -9, -4>| / |<7, -9, -4>|
= |<29/29, 0, 0 - 95/29> · <7, -9, -4>| / |<7, -9, -4>|
= |<29/29, 0, -95/29> · <7, -9, -4>| / |<7, -9, -4>|
= |<203/29, 0, -380/29>| / |<7, -9, -4>|
= (√[(203/29)^2 + 0 + (-380/29)^2]) / (√(7^2 + (-9)^2 + (-4)^2))
= (√(41209/841 + 144400/841)) / (√146)
= (√(185609/841)) / (√146)
= (√185609) / (√(841 * 146))
= (√185609) / (√122486)
= (√185609) / 349.969

Therefore, the distance between the skew lines is (√185609) / 349.969 (approximately).