Equations of a line in space

What you will be able to do

• write a line through a point and parallel to a vector, in vector and in Cartesian form;
• write the line through two points;
• move between the two forms and test whether a point lies on a line.

A line in space

A kite string pulled tight is a straight line. Hold one end at a point $A$ and let the string point along a direction $\vec{b}$. Every point $P$ of the string is reached by walking from $A$ some multiple of $\vec{b}$. In symbols, with $\vec{a}$ the position vector of $A$ and $\vec{r}$ the position vector of $P$:

The number $\lambda$ (lambda) is the parameter. Each real value of $\lambda$ gives one point of the line. A-Level and IB write $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$, or $\mathbf{r} = \mathbf{a} + t\mathbf{b}$. They are all the same equation.

The Cartesian form

Put $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. Then $x = x_1 + \lambda a$, $y = y_1 + \lambda b$ and $z = z_1 + \lambda c$. Solve each for $\lambda$ and set them equal:

This is the symmetric form of the line. The denominators $a$, $b$, $c$ are direction ratios. If they are replaced by the direction cosines $l$, $m$, $n$, the common value is the actual distance from $A$ along the line.

When a denominator is $0$, the matching numerator is $0$ as well. For example $\dfrac{x - 1}{3} = \dfrac{y + 2}{0} = \dfrac{z - 4}{-2}$ means that $y = -2$ all along the line, and $\dfrac{x - 1}{3} = \dfrac{z - 4}{-2}$. The fraction with $0$ underneath is only a way of writing "$y$ does not change".

The line through two points

If the line passes through $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, its direction is $\vec{AB}$, and

Here $\vec{b}$ is the position vector of $B$. Notice the letter $\vec{b}$ now means a point, and not a direction. Say which one you mean when you write.

A worked example

A line passes through $A(1, 2, 3)$ and is parallel to $\vec{b} = 2\hat{i} - \hat{j} + 4\hat{k}$.

1. Vector form. $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - \hat{j} + 4\hat{k})$.
2. Cartesian form. $\dfrac{x - 1}{2} = \dfrac{y - 2}{-1} = \dfrac{z - 3}{4}$.
3. Does $(7, -1, 15)$ lie on it? Each fraction gives $\dfrac{7 - 1}{2} = 3$, $\dfrac{-1 - 2}{-1} = 3$ and $\dfrac{15 - 3}{4} = 3$. They agree, so yes, with $\lambda = 3$.
4. Does $(3, 1, 8)$ lie on it? The fractions are $1$, $1$ and $\dfrac{5}{4}$. They disagree, so no.

A line through two points. Take $A(2, -1, 4)$ and $B(5, 3, -8)$. The direction ratios are $(3, 4, -12)$, the length is $13$, and the line is

From Cartesian to vector form. The line $\dfrac{x - 3}{2} = \dfrac{y + 1}{3} = \dfrac{z - 5}{6}$ passes through $(3, -1, 5)$ and has direction ratios $(2, 3, 6)$. So $\vec{r} = (3\hat{i} - \hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$. Read the signs carefully: $y + 1$ means $y_1 = -1$.

Practise with new numbers

Try it in Python

The program lists points of the line of the worked example for several values of $\lambda$. It then tests whether $(7, -1, 15)$ lies on the line.

Common mistakes

• Wrong sign of the point. In $\dfrac{y + 1}{3}$ the coordinate is $y_1 = -1$. Move the number across before you read it.
• Using the point as the direction. The denominators are direction ratios, not the coordinates of the point.
• Testing only two coordinates. All three fractions must agree.
• Forgetting the parameter. Different values of $\lambda$ give different points, and one value gives one point.

Remember these

• Vector form: $\vec{r} = \vec{a} + \lambda\vec{b}$. The line passes through $\vec{a}$ and is parallel to $\vec{b}$.
• Cartesian form: $\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$.
• Through two points: use the differences of the coordinates as direction ratios.
• A zero denominator means that coordinate stays fixed along the line.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write the vector equation of the line through $(1, 0, -2)$ parallel to $3\hat{i} + \hat{j} - 4\hat{k}$.

2. Write $\dfrac{x + 2}{3} = \dfrac{y - 1}{2} = \dfrac{z - 4}{-1}$ in vector form.

3. Find the line through $(1, 1, 1)$ and $(3, 2, -1)$ in Cartesian form.

4. Does $(4, 3, -4)$ lie on the line of question 3?

5. What does $\dfrac{x - 5}{0} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2}$ say about $x$?

References

• OpenStax, Calculus Volume 3, Chapter 2, section on equations of lines and planes in space (vector, parametric and symmetric equations of a line).
• NCERT, Mathematics Part II for Class XII, Chapter 11, section on the equation of a line in space.