Skew lines

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In geometry, two lines are said to be skew lines if they do not intersect but are not parallel.

Skew lines only exist in three or more dimensions; any two distinct lines in the plane which are not parallel must intersect at some point. In fact, two lines are skew lines if and only if they do not lie in a single plane together. This means that if each line is defined by two points, these four points must not be coplanar; put another way, they must be vertices of a tetrahedron of nonzero volume. Any three of them will still be coplanar, since three points define a plane, but no three points will be collinear, since this would make all four points coplanar.

The volume of a tetrahedron with four noncoplanar vertices v₁, v₂,v₃ and v₄ in n-dimensional space Euclidean space Rⁿ can be found using the wedge product; it is one-sixth of the quantity

[||(v_2-v_1) \wedge (v_3-v_1) \wedge (v_4-v_1)|| ]

which defines the volume of a corresponding parallelepiped. Here ||*|| denotes the norm; that is, the square root of the sum of the squares of the coefficients of the wedge product. Since this volume must be nonzero for the four points to define skew lines, we can identify skew lines as those satisfying the condition that the above wedge product is not zero.

In three dimensions, if the verticies are written v₁=(x₁,y₁,z₁), v₂=(x₂,y₂,z₂), v₃=(x₃,y₃,z₃), and v₄=(x₄,y₄,z₄) then this volume is the absolute value of the determinant

[\begin x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end \ne 0.]

Two randomly chosen lines will usually be a skew line, because given any three points defining a plane, that plane has zero volume, and so there is zero probability that a randomly-chosen fourth point will fall on it (the probability that the first three points will not define a plane is also zero). Similarly, a very small perturbation of two parallel or intersecting lines will almost surely turn them into skew lines. In this sense, skew lines are the "usual" case, and parallel or intersecting lines are special cases.

Formulas involving lines

If u is a vector defining a point, and v₁ and v₂ determine the line t(v₂-v₁) - v₁, then by taking dot products the square of the distance is [At^2 + 2Bt + C,] where [A = (v_2-v_1) \cdot (v_2-v_1), B = 2(v_2-v_1) \cdot (v_1-u),] [C=(v_1-u) \cdot (v_1-u).] From this we may find the minimum distance as

[d^2 = \frac.]

Using the Lagrange identity, we may verify this expression is equivalent to

[d = \frac

>.]

If now two lines are given by t(v₂-v₁) - v₁ and s(v₄-v₃) - v₃, we may again use the dot product to find the square of the distance is

[As^2+2Bst+Ct^2+2Ds+2Et+F,]

where

[A = (v_4-v_3) \cdot (v_4-v_3), B=(v_4-v_3) \cdot (v_1-v_2),]

[C = (v_1-v_2) \cdot (v_1-v_2), D=(v_4-v_3) \cdot (v_3-v_1),]

[E=(v_1-v_2) \cdot (v_3-v_1), F=(v_3-v_1) \cdot (v_3-v_1).]

Finding the minimum of this expression, we obtain the minimum distance between two lines as

[d^2 = \frac = \frac]

where

[S=\beginA&B\\B&C\end]

and det R is the Hessian, where

[R=\beginA&B&D\\B&C&E\\D&E&F\end.]

Suppose a, b, and c are any vectors in n-space, and N is the vector with rows a, b, and c. If

[M = N^T N = \begina \cdot a&a \cdot b&a \cdot c\\b \cdot a&b \cdot b&b \cdot c\\c \cdot a&c \cdot b&c \cdot c\end]

then using the generalized Lagrange identity,

[(a \wedge b \wedge c) \cdot (a \wedge b \wedge c) = \det M]

we arrive at the elegant formula

[d = \frac

>.]

This may be written

[d = \frac

>,]

and so the numerator is the volume of the parallelepiped defined by v₁, v₂, v₃ and v₄.

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