# Skew lines and Plane Intersectionsline 1 is the lines a which 3 planes intersect `x-y+z=1` `y+z=2` `2x+4z=6` line 2 is `x-1/2=y=z+2/3` Show that line 1 and line 2 are skewed.

txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

First, let's parameterize Line 2 to simplify it and make it into a vector relationship. Let's let `G(x(t),y(t),z(t))` be line 2 and `y=t`. We can now get a parametric relationship:

`x = t+1/2`

`y = t`

`z=t-2/3`

We can write G in vector form as follows:

`vecG(x,y,z) = (t+1/2)veci + tvecj + (t-2/3)veck`

Now, let's focus on finding line 1. Because we know all three planes intersect in a line, their equations are not linearly independent. Therefore, we just need to take 2 of the equations and parameterize to get an equation for their line of intersection (we'll call it `F(x,y,z)`). Let's also let `z = t`. For simplicity, let's use the two equations that only have 2 variables:

`y + z = 2`

`2x+4z = 6`

If we solve for x and y in terms of z (in our case, this is solving in terms of t, too!):

`y = 2-z=2-t`

`x = 3-2z = 3-2t`

We can now get our parameterized line:

`x = 3-2t`

`y = 2-t`

`z = t`

Written in vector form,

`vecF(x,y,z) = (3-2t)veci + (2-t)vecj + tveck`

Now, the best way to show that lines are skew is to show that their x components, y components, and z components cannot be equal and that their 3-D "slopes" are different.

First, let's show that they never intersect. Let `t_1` be the parameter for F and let `t_2` be the parameter for G. If x, y, and z are equal between F and G, the following equations should show linear dependence (in other words, one of the 3 equations can be expressed as a linear combination of the other 2). I'm not sure if you've taken linear algebra yet, so I'll just explicitly show there is no solution. We equate each value of x, y, and z:

`3-2t_1 = t_2 + 1/2`

`2-t_1=t_2`

`t_1 = t_2-2/3`

We don't need to go far to disprove the idea there may be an intersection. Look at the last two equations when you solve for `t_1` in terms of `t_2` :

`t_1 = 2-t_2`

`t_1 = t_2 - 2/3`

Clearly these equations have no solution, so the lines do not intersect.

Now, we show that the lines are not parallel (implying they are skew). The 3-D "slope" of the lines is similar to the slope of a 2-D line on the x-y axes. We simply look at what we're multiplying by t in each parameterized equation (or vector form). If you've had calculus, it means that the derivatives `(dF)/dt` and `(dG)/dt` must be be related by a constant proportion. In other words, there must be some constant C such that:

`(dF)/dt = C(dG)/dt`

Let's look at another vector form of our lines:

`vecF(x,y,z) = (3,2,0) + t(-2,-2,1)`

`vecG(x,y,z) = (1/2, 0, -2/3) + t(1,1,1)`

In order for these lines to be skew, the following must hold true:

`(-2,-1,1) = C(1,1,1)`

Clearly there is no constant C that can be multiplied by 1 to get both -2 and -1. Therefore, these lines are not parallel.

Because the lines do not intersect and are not parallel, they must be skew.