Angle between two lines & Parallel, intersecting, skew lines

Acute angle between two lines

The acute angle between 2 lines can be found by $$ \boxed{ \cos \theta = { | \textbf{m}_1 \cdot \textbf{m}_2 | \over | \textbf{m}_1 | | \textbf{m}_2| } } $$

$ \textbf{m}_1$ and $ \textbf{m}_2$ are the direction vectors of each line.

Visualising parallel, intersecting, skew lines

In the triangular prism above, the parallel lines are

• Lines AD and BC
• Lines ED and FC
• Lines AE and BF
• Lines AB, DC and EF

Lines that are not parallel and do not intersect are considered skew lines. For example, AD and CF are skew lines.

In short, to determine if two lines are skew lines:

1. Check whether lines are parallel
2. If lines are not parallel, check whether lines have a common point of intersection
3. If lines do not have a common point of intersection, lines are skew lines

Check whether lines are parallel, intersecting or skew

\begin{align} l_1 : \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right), \lambda \in \mathbb{R} \\ \\ l_2 : \textbf{r} & = \mu \left( \begin{matrix} -1 \\ 1.5 \\ -0.5 \end{matrix} \right), \mu \in \mathbb{R} \end{align}

Check whether the direction vectors are parallel: \begin{align} \text{Since } \left( \begin{matrix} 2 \\ - 3 \\ 1 \end{matrix} \right) & = -2 \left( \begin{matrix} -1 \\ 1.5 \\ -0.5 \end{matrix} \right) , \phantom{0} l_1 \text{ is parallel to } l_2 \end{align}

\begin{align} l_1 : \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right), \lambda \in \mathbb{R} \\ \\ l_3 : \textbf{r} & = \left( \begin{matrix} -1 \\ 1 \\ 3 \end{matrix} \right) + \alpha \left( \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right), \alpha \in \mathbb{R} \end{align}

\begin{align} \text{Let } \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) & = \left( \begin{matrix} -1 \\ 1 \\ 3 \end{matrix} \right) + \alpha \left( \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} 1 + 2\lambda \\ 1 -3\lambda \\ 1 + \lambda \end{matrix} \right) & = \left( \begin{matrix} -1 + \alpha \\ 1 - 2\alpha \\ 3 + \alpha \end{matrix} \right) \\ \\ 1 + 2\lambda & = -1 + \alpha \implies 2 \lambda - \alpha = - 2 \phantom{0} \text{--- (1)} \\ 1 - 3\lambda & = 1 - 2\alpha \implies - 3 \lambda + 2\alpha = 0 \phantom{0} \text{--- (2)} \\ 1 + \lambda & = 3 + \alpha \implies \lambda - \alpha = 2 \phantom{0} \text{--- (3)} \\ \\ \text{Solve } (1) \& & (2) \text{ by GC, } \lambda = -4 \text{ and } \alpha = -6 \\ \\ \text{Substitute } \lambda = -4 & \text{ and } \alpha = - 6 \text{ into (3),} \\ \text{L.H.S} & = (-4) - (-6) \\ & = 2 \\ & = \text{R.H.S} \\ \\ \text{Substitute } & \lambda = - 4 \text{ into } l_1, \\ \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + (-4) \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \\ & = \left( \begin{matrix} -7 \\ 13 \\ -3 \end{matrix} \right) \\ \\ \text{Point of intersection} & \text{ between } l_1 \text{ and } l_3 \text{ is } (-7, 13, - 3) \end{align}

\begin{align} l_1 : \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right), \lambda \in \mathbb{R} \\ \\ l_4 : \textbf{r} & = \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) + \beta \left( \begin{matrix} 1 \\ 1 \\ -1 \end{matrix} \right), \beta \in \mathbb{R} \end{align}

\begin{align} \text{Let } \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) & = \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) + \beta \left( \begin{matrix} 1 \\ 1 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 1 + 2\lambda \\ 1 -3\lambda \\ 1 + \lambda \end{matrix} \right) & = \left( \begin{matrix} \beta \\ \beta \\ 1 - \beta \end{matrix} \right) \\ \\ 1 + 2\lambda & = \beta \implies 2\lambda - \beta = -1 \phantom{0} \text{--- (1)} \\ 1 - 3\lambda & = \beta \implies -3\lambda - \beta = -1 \phantom{0} \text{--- (2)} \\ 1 + \lambda & = 1 - \beta \implies \lambda + \beta = 0 \phantom{0} \text{--- (3)} \\ \\ \text{Solve } (1) \& & (2) \text{ by GC, } \lambda = 0 \text{ and } \beta = 1 \\ \\ \text{Substitute } \lambda = 0 & \text{ and } \beta = 1 \text{ into (3),} \\ \text{L.H.S} & = 0 + 1 \\ & = 1 \ne \text{R.H.S} \\ \\ \implies \text{No common point} & \text{ of intersection between } l_1 \text{ and } l_4 \\ \implies l_1 \text{ and } l_4 \text{ are skew } & \text{lines} \end{align}