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Vectors – The vector equation of a line

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Vectors — The Vector Equation of a Line

In three dimensions, a line can be described using vectors. This is very useful in Year 13 because it gives a clear way to describe the position of any point on the line.

A vector equation of a line needs a fixed point on the line and a direction vector. From this, we can write the line in column vector form, parametric form, and \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form.

Key definitions and formulae

1. Vector equation of a line in column vector form

If a line passes through the point \( A(x_1,y_1,z_1) \) and has direction vector

\[ \begin{pmatrix} a\\ b\\ c \end{pmatrix}, \]

then the vector equation of the line is

\[ \mathbf{r}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} + \lambda \begin{pmatrix} a\\ b\\ c \end{pmatrix}, \]

where \( \lambda \) is a parameter.

2. Meaning of the parameter

The parameter \( \lambda \) can take any real value. Different values of \( \lambda \) give different points on the line.

\[ \lambda \in \mathbb{R}. \]

3. Parametric form

From

\[ \mathbf{r}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} + \lambda \begin{pmatrix} a\\ b\\ c \end{pmatrix}, \]

we get the parametric equations

\[ x=x_1+a\lambda, \qquad y=y_1+b\lambda, \qquad z=z_1+c\lambda. \]

4. \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form

The same line can also be written as

\[ \mathbf{r}=(x_1\mathbf{i}+y_1\mathbf{j}+z_1\mathbf{k}) + \lambda(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}). \]

This is the same equation written using unit vectors.

5. Important ideas

Part of equation Meaning
\[ \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} \] A fixed point on the line
\[ \begin{pmatrix} a\\ b\\ c \end{pmatrix} \] A direction vector of the line
\( \lambda \) A parameter that generates all points on the line

Worked examples

Example 1: Write the equation of a line through a point with a given direction vector

Write the vector equation of the line through \( A(2,-1,3) \) with direction vector

\[ \begin{pmatrix} 4\\ -2\\ 1 \end{pmatrix}. \]

Solution

Using

\[ \mathbf{r}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} + \lambda \begin{pmatrix} a\\ b\\ c \end{pmatrix}, \]

we get

\[ \mathbf{r}= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4\\ -2\\ 1 \end{pmatrix}. \]

Parametric form:

\[ x=2+4\lambda, \qquad y=-1-2\lambda, \qquad z=3+\lambda. \]

\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:

\[ \mathbf{r}=(2\mathbf{i}-\mathbf{j}+3\mathbf{k}) + \lambda(4\mathbf{i}-2\mathbf{j}+\mathbf{k}). \]
All three forms describe the same line.

Example 2: Find the vector equation of the line through two points

Find the equation of the line through \( A(1,2,-3) \) and \( B(5,-1,7) \).

Solution

First find a direction vector:

\[ \overrightarrow{AB}= \begin{pmatrix} 5-1\\ -1-2\\ 7-(-3) \end{pmatrix} = \begin{pmatrix} 4\\ -3\\ 10 \end{pmatrix}. \]

Now use point \( A \):

\[ \mathbf{r}= \begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 4\\ -3\\ 10 \end{pmatrix}. \]

Parametric form:

\[ x=1+4\lambda, \qquad y=2-3\lambda, \qquad z=-3+10\lambda. \]

\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:

\[ \mathbf{r}=(\mathbf{i}+2\mathbf{j}-3\mathbf{k}) + \lambda(4\mathbf{i}-3\mathbf{j}+10\mathbf{k}). \]
A line through two points is found by using one point and the displacement vector between the points.

Example 3: Find whether a point lies on a line

Determine whether the point \( P(7,-4,5) \) lies on the line

\[ \mathbf{r}= \begin{pmatrix} 1\\ 2\\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2\\ -2\\ 3 \end{pmatrix}. \]

Solution

From the vector equation,

\[ x=1+2\lambda, \qquad y=2-2\lambda, \qquad z=-1+3\lambda. \]

Using \( x=7 \):

\[ 1+2\lambda=7 \qquad \Rightarrow \qquad 2\lambda=6 \qquad \Rightarrow \qquad \lambda=3. \]

Check the other coordinates:

\[ y=2-2(3)=-4, \qquad z=-1+3(3)=8. \]

The \( y \)-coordinate matches, but the \( z \)-coordinate does not match \( 5 \).

So \( P(7,-4,5) \) does not lie on the line.

Example 4: Find the point where \( \lambda=2 \)

For the line

\[ \mathbf{r}= \begin{pmatrix} -3\\ 1\\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 5\\ 2\\ -1 \end{pmatrix}, \]

find the coordinates of the point when \( \lambda=2 \).

Solution

\[ \mathbf{r}= \begin{pmatrix} -3\\ 1\\ 4 \end{pmatrix} + 2 \begin{pmatrix} 5\\ 2\\ -1 \end{pmatrix} = \begin{pmatrix} -3\\ 1\\ 4 \end{pmatrix} + \begin{pmatrix} 10\\ 4\\ -2 \end{pmatrix} = \begin{pmatrix} 7\\ 5\\ 2 \end{pmatrix}. \]
The point is \((7,5,2)\).

Example 5: Convert from \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form to parametric form

Convert

\[ \mathbf{r}=(3\mathbf{i}-2\mathbf{j}+\mathbf{k}) + t(2\mathbf{i}+5\mathbf{j}-4\mathbf{k}) \]

into parametric equations.

Solution

Match the coefficients of \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \):

\[ x=3+2t, \qquad y=-2+5t, \qquad z=1-4t. \]
The parametric form is \[ x=3+2t,\qquad y=-2+5t,\qquad z=1-4t. \]

Standard methods

Method 1: Write the equation of a line from a point and a direction vector

  1. Write the coordinates of the point as a position vector.
  2. Write the direction vector.
  3. Substitute into \[ \mathbf{r}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} + \lambda \begin{pmatrix} a\\ b\\ c \end{pmatrix}. \]

Method 2: Write the equation of a line through two points

  1. Find a direction vector by subtracting the coordinates: \[ \overrightarrow{AB}= \begin{pmatrix} x_2-x_1\\ y_2-y_1\\ z_2-z_1 \end{pmatrix}. \]
  2. Use either point as the fixed point.
  3. Substitute into the vector equation of the line.

Method 3: Check whether a point lies on a line

  1. Write the parametric equations.
  2. Use one coordinate to find the parameter.
  3. Substitute into the other coordinates to check consistency.

Common mistakes and exam tips

Common mistakes

  • Using a point that is not on the line as the fixed position vector.
  • Finding the direction vector in the wrong order when using two points.
  • Forgetting that any non-zero multiple of a direction vector gives the same line.
  • Confusing the vector equation with the parametric equations.
  • When checking a point, using one coordinate only and not checking the others.

Exam tips

  • Always identify clearly: point on the line and direction vector.
  • Write all three forms when asked: column vector form, parametric form, and \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form.
  • If a direction vector can be simplified, that is fine, but it is not essential unless the question asks for it.
  • When using a parameter, keep the same letter throughout your working.

Summary

\[ \mathbf{r}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} + \lambda \begin{pmatrix} a\\ b\\ c \end{pmatrix} \] \[ x=x_1+a\lambda,\qquad y=y_1+b\lambda,\qquad z=z_1+c\lambda \] \[ \mathbf{r}=(x_1\mathbf{i}+y_1\mathbf{j}+z_1\mathbf{k}) + \lambda(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}) \]

The vector equation of a line describes every point on the line using one fixed point and one direction vector. It can be written in column vector form, parametric form, or \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form.