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Lines

Level 1 - Math II (Physics) topic page in Vectors.

Principle

A straight line is a set of points that all lie in one fixed direction from a chosen reference point. In vector form, a line in two-dimensional space or three-dimensional space is described by a point on the line and a non-zero direction vector parallel to the line.

This vector-line description works in \(\mathbb R^2\), where vectors have two components, and in \(\mathbb R^3\), where vectors have three components.

Notation

\(\ell\)
a straight line
\(A\)
a fixed reference point on the line
\(\mathbf a\)
the position vector of A from the chosen origin
\(P\)
a variable point on the line
\(\mathbf r\)
the position vector of P from the chosen origin
\(\mathbf b\)
a non-zero direction vector parallel to the line
\(\lambda\)
a real parameter that selects a point on the line
\(\mathbf c\)
the position vector of a point C whose distance from the line is being measured
\(\mathbf a-\mathbf c\)
a vector from C to the reference point A
\((\mathbf a-\mathbf c)\times\mathbf b\)
a vector whose magnitude gives the area of the parallelogram formed by \mathbf a-\mathbf c and \mathbf b

The closest-distance vector from a point \(C\) to a line is the vector from \(C\) to the nearest point on \(\ell\). It is perpendicular to the line, so it is perpendicular to every direction vector for the line.

Method

Step 1: Choose one point on the line

Choose any known point \(A\) on \(\ell\). Write its position vector as \(\mathbf a\). In \(\mathbb R^2\), \(\mathbf a=(a_1,a_2)\). In \(\mathbb R^3\), \(\mathbf a=(a_1,a_2,a_3)\).

Step 2: Choose a non-zero parallel vector

Choose any non-zero vector \(\mathbf b\) parallel to the line. The vector \(\mathbf b\) must not be \(\mathbf 0\), because \(\lambda\mathbf 0=\mathbf 0\) for every \(\lambda\), so it would not create different points along a line.

Step 3: Write the line equation

Start at the reference position \(\mathbf a\), then add a variable multiple of the direction vector:

Line from point and direction
\[\mathbf r=\mathbf a+\lambda\mathbf b\]

The parameter \(\lambda\) runs through all real numbers. Positive values move in the direction of \(\mathbf b\), negative values move in the opposite direction, and \(\lambda=0\) gives the reference point \(A\).

Rules

Line equation
\[\ell:\quad\mathbf r=\mathbf a+\lambda\mathbf b,\quad \lambda\in\mathbb R,\quad \mathbf b\ne\mathbf 0\]
Rescaled direction
\[\mathbf r=\mathbf a+t(k\mathbf b),\quad k\ne0,\quad t\in\mathbb R\]
Changed reference point
\[\mathbf a'=\mathbf a+\lambda_0\mathbf b\quad\Longrightarrow\quad \mathbf r=\mathbf a'+t\mathbf b\]
Distance from a point to a line
\[d=\frac{|(\mathbf a-\mathbf c)\times\mathbf b|}{|\mathbf b|}\]

The distance formula applies in \(\mathbb R^3\). For a line in the \(xy\)-plane, the same formula can be used by treating all vectors as three-dimensional vectors with zero \(z\)-component.

Why \(\mathbf r-\mathbf a\) is parallel to \(\mathbf b\)

If \(P\) is on the line, then its position vector satisfies \(\mathbf r=\mathbf a+\lambda\mathbf b\) for some real \(\lambda\). Subtract \(\mathbf a\) from both sides:

Start with the line equation
\[\mathbf r=\mathbf a+\lambda\mathbf b\]
Subtract \mathbf a from both sides
\[\mathbf r-\mathbf a=(\mathbf a+\lambda\mathbf b)-\mathbf a\]
Regroup the vector sum
\[\mathbf r-\mathbf a=(\mathbf a-\mathbf a)+\lambda\mathbf b\]
Use \mathbf a-\mathbf a=\mathbf 0
\[\mathbf r-\mathbf a=\mathbf 0+\lambda\mathbf b\]
Use \mathbf 0+\lambda\mathbf b=\lambda\mathbf b
\[\mathbf r-\mathbf a=\lambda\mathbf b\]

Because \(\mathbf r-\mathbf a\) is a scalar multiple of \(\mathbf b\), it is parallel to \(\mathbf b\). When \(\lambda=0\), the vector \(\mathbf r-\mathbf a\) is \(\mathbf 0\), which represents the reference point itself.

Why the distance formula works

Let \(C\) have position vector \(\mathbf c\). The vector from \(C\) to \(A\) is \(\mathbf a-\mathbf c\). If \(d\) is the perpendicular distance from \(C\) to the line, then the parallelogram with side vectors \(\mathbf a-\mathbf c\) and \(\mathbf b\) has base length \(|\mathbf b|\) and height \(d\).

Area from base and height
\[\text{area}=|\mathbf b|d\]
Area from cross product
\[\text{area}=|(\mathbf a-\mathbf c)\times\mathbf b|\]
Equate the two area expressions
\[|\mathbf b|d=|(\mathbf a-\mathbf c)\times\mathbf b|\]
Divide by the non-zero base length
\[d=\frac{|(\mathbf a-\mathbf c)\times\mathbf b|}{|\mathbf b|}\]

Examples

Question
Write a vector equation for the line in the \(xy\)-plane with
\[x=1\]
parallel to the \(y\)-axis.
Answer
Every point on the line has
\[x=1\]
Choose
\[A=(1,0)\]
so
\[\mathbf a=\mathbf i\]
A non-zero vector parallel to the \(y\)-axis is
\[\mathbf b=\mathbf j\]
Therefore
\[\mathbf r=\mathbf i+\lambda\mathbf j\]
with
\[\lambda\in\mathbb R\]
In coordinates this is
\[\mathbf r=(1,0)+\lambda(0,1)=(1,\lambda)\]
so the \(x\)-component is always
\[1\]

Checks

  • The direction vector must not be the zero vector; otherwise the equation gives only one point.
  • Many parametrizations can describe the same line because direction vectors can be rescaled and reference points can be changed to any point on the line.
  • The parameter has no fixed unit unless the chosen direction-vector convention gives it one.

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