# Show that the following lines are skew and nd the shortest distance between them. L_1: x = 1 -2t, y = 1 + t, z = 3t L_2: (x+ 2)/2= y = z -1

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### 1 Answer

You need to prove that the lines `L_1` and ` L_2` are skew, hence, you need to prove that the lines are not parallel, nor they intersect.

You need to convert the original form of equation of line `L_2` into paramateric form, such that:

`L_2: {(x + 2 = 2s),(y = s),(z - 1 = s):} => {(x = -2 + 2s),(y = s),(z = 1 + s):}`

You need to check if the direction vectors of `L_1` and `L_2` are proportional, such that:

`L_1: <-2,1,3>`

`L_2: <2,1,1>`

You should notice that proportions -`2/2 != 1/1 != 3/1` , hence, the lines are not parallel.

You need to test if the lines intersect, hence, you need to solves the following system of equations, such that:

`{(x = 1 - 2t = -2 + 2s),(y = 1 + t = s),(z = 3t = 1 + s):}`

Solving the system of equations `{(2s + 2t = 3),(s = 1 + t):}` yields:

`2(1 + t) + 2t = 3 => 2 + 2t + 2t = 3 => 4t = 3 - 2 => t = 1/4 => s = 1 + 1/4 => s = 5/4`

Solving the system of equations `{(s - 3t = -1),(s = 1 + t):}` yields:

`1 + t - 3t = -1 => -2t = -2 => t = 1 => s = 2`

Since the values obtained for t and s, soving the systems of equations above, do not coincide, the lines do not intersect.

**Hence, since the lines are not parallel, nor they intersect, yields that `L_1` and `L_2 ` are skew lines.**

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