# Relative position of two straight lines

The implicit equations of the straight lines and the plane are linear equations. For this reason, we can determine the relative position of the straight lines and the planes in space from the study of the compatibility of the systems of corresponding linear equations.

Let's see the relative positions that can have straight lines, $$r (A;\overrightarrow{v})$$ and $$r'(A';\overrightarrow{v'})$$, given by their respective implicit equations:

$$r:\left\{ \begin{array}{rcl} A_1x+B_1y+C_1z+D_1&=&0 \\ A_2x+B_2y+C_2z+D_2&=&0 \end{array}\right. \\ r':\left\{ \begin{array}{rcl} A_3x+B_3y+C_3z+D_3&=&0 \\ A_4x+B_4y+C_4z+D_4&=&0 \end{array}\right.$$$To find the relative positions between the straight lines, let's consider the system formed by four equations, whose matrixes are $$M$$ and $$M'$$: $$M=\begin{pmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 &C_3 \\ A_4 & B_4 & C_4\end{pmatrix} \qquad M'=\begin{pmatrix} A_1 & B_1 & C_1 & -D_1 \\ A_2 & B_2 & C_2 & -D_2 \\ A_3 & B_3 &C_3 & -D_3 \\ A_4 & B_4 & C_4 & -D_4\end{pmatrix}$$$

Using the ranks of the above mentioned matrixes, we will determine the compatibility of the systems, and with that, the relative position of the straight lines:

## Compatible system

### Determined

$$rank(M) = rank(M') = 3$$

Determined Compatible system. The straight lines are secant, that is, they cut at a point.

### Indeterminate

$$rank(M) = rank (M') = 2$$

Indeterminate compatible system which solutions depend on a parameter. The straight lines are the same.

## Incompatible system:

$$rank (M) = 2$$; $$rank (M') = 3$$

Incompatible system. Two straight lines are parallel.

$$rank (M) = 3$$; $$rank (M') = 4$$

Incompatible system. Two straight lines cross.

Determine the relative position of the following straight lines:

$$r:\left\{ \begin{array}{rcl} 7x+5y-7z-12 & = & 0 \\2x+3z+11 & = & 0 \end{array}\right. \qquad r':\left\{ \begin{array}{rcl} 5x-5y-z-16 & = & 0\\3x-2y-7 & = & 0 \end{array}\right.$$$The matrixes of the system formed by four equations are: $$M=\begin{pmatrix} 7 & 5 & -7\\ 2 & 0 & 3 \\ 5& -5 &-1\\ 3& -2 &0\end{pmatrix} \qquad M'=\begin{pmatrix} 7&5&-7 &12\\ 2 & 0 & 3 &-11\\ 5& -5 &-1&16 \\ 3& -2 & 0 &7 \end{pmatrix}$$$

And if we calculate the rank of the above mentioned matrixes we have:

$$\left|\begin{matrix}7&5&-7\\ 2 & 0 & 3 \\ 5& -5 &-1\\ 3& -2 &0\end{matrix}\right|=260 \neq 0 \Rightarrow \mbox{ rank }(M)=3 \\ \\ \left|\begin{matrix}7&5&-7 &12\\ 2 & 0 & 3 &-11\\ 5& -5 &-1&16 \\ 3& -2 & 0 &7 \end{matrix}\right|=552 \neq 0 \Rightarrow \mbox{ rank }(M')=4$$\$

Therefore the straight lines cross.

The relative position of two straight lines can also be determined from a more geometric and not such an algebraic point of view from its respective director vectors.

## Linearly independent director vectors: The straight lines either cut or cross

To distinguish both cases, we will create a third vector from a point of the first straight line to another point of another straight line. If the above mentioned vector is linearly dependent with the director vectors of the straight lines, then they cut. If not, they cross.

## Linearly dependent director vectors: The straight lines are parallel or equal

To distinguish both cases we will take a point in one of the two lines and we will check if it also belongs to the other line. If it does, then the two lines are the same. If not, they are parallel.