# Angle between lines & Parallel, intersecting, skew lines

Parallel and Skew Lines
Parallel and Skew Lines

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

Can you identify the parallel lines, intersecting lines and skew lines in the triangular prism above?

In the triangular prism above, the parallel lines are

• 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

Case of parallel lines

\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}

Case of intersecting lines

\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}

From observation, it is clear that both direction vectors are not parallel.

Check whether there is a point of intersection between the two lines:

\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}

Case of skew lines

\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}

From observation, it is clear that both direction vectors are not parallel.

Check whether there is a point of intersection between the two lines:

\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}

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