The Cartesian Form of a Line

Three equations sharing one t

The vector equation r = (1, 2, 3) + t(2, 1, 4) says three things at once, one for each coordinate. Read it row by row: x = 1 + 2t, y = 2 + t and z = 3 + 4t.

All three rows use the same t. Choosing t = 2 gives the point (5, 4, 11), and each coordinate of that point comes from the one value t = 2.

Make t the subject

Rearrange each row to say what t is. From x = 1 + 2t, take 1 from both sides and divide by 2: $t = \frac{{x - 1}}{2}$. From y = 2 + t: $t = \frac{{y - 2}}{1}$. From z = 3 + 4t: $t = \frac{{z - 3}}{4}$.

At any point of the line, these three expressions are all equal to the same number, that point’s t.

Set them equal

Since all three equal t, they equal each other: $\frac{{x - 1}}{2} = \frac{{y - 2}}{1} = \frac{{z - 3}}{4}$. This is the Cartesian form of the line. The parameter has gone, and what is left is two equations linking x, y and z.

A point lies on the line exactly when its three fractions agree. For (5, 4, 11): $\frac{{5 - 1}}{2} = 2$, $\frac{{4 - 2}}{1} = 2$ and $\frac{{11 - 3}}{4} = 2$, all equal, so the point is on the line, at t = 2. For (5, 4, 10) the third fraction is $\frac{7}{4}$, not 2, so that point is not on the line.

The flat line $r = \left(1, 1\right) + \lambda\left(2, 1\right)$, whose Cartesian form is $\frac{{x - 1}}{2} = \frac{{y - 1}}{1}$. At $\lambda = 1.5$ the point is (4, 2.5), and $\frac{{4 - 1}}{2} = 1.5$ and $\frac{{2.5 - 1}}{1} = 1.5$. Drag the point: both fractions stay equal to $\lambda$ wherever it goes.

Reading it back

In general, the line through the point $\left(a_{1}, a_{2}, a_{3}\right)$ with direction $\left(d_{1}, d_{2}, d_{3}\right)$ is $\frac{{x - a_{1}}}{d_{1}} = \frac{{y - a_{2}}}{d_{2}} = \frac{{z - a_{3}}}{d_{3}}$. The numbers taken away on top give the point; the numbers underneath give the direction.

Watch the signs. $\frac{{x + 2}}{3} = \frac{{y - 1}}{{-2}} = \frac{z}{5}$ is $\frac{{x - \left(-2\right)}}{3} = \frac{{y - 1}}{{-2}} = \frac{{z - 0}}{5}$, so the point is (−2, 1, 0) and the direction is (3, −2, 5).

Put it in the standard form first

The reading only works when each top is x, y or z minus a number, with nothing multiplying the letter. Take $\frac{{2x - 4}}{6} = \frac{{y + 1}}{3} = \frac{{3 - z}}{2}$.

The first fraction is $\frac{2\left(x - 2\right)}{6} = \frac{{x - 2}}{3}$. The third is $\frac{{3 - z}}{2} = - \frac{{z - 3}}{2} = \frac{{z - 3}}{{-2}}$. So the line is $\frac{{x - 2}}{3} = \frac{{y + 1}}{3} = \frac{{z - 3}}{{-2}}$: the point (2, −1, 3) and the direction (3, 3, −2).

A zero in the direction

Take r = (5, 2, 1) + t(3, 0, 4). The middle row is y = 2 + 0t, so y = 2 at every point of the line, and t cannot be made the subject of that row: there is nothing to divide by.

So that coordinate is written on its own: $\frac{{x - 5}}{3} = \frac{{z - 1}}{4}$, y = 2. The line lies in the plane y = 2.

With two zeros, two coordinates are fixed. The line through (1, 2, 0) with direction (0, 0, 1) is x = 1, y = 2: a vertical line, with z free.

The plane y = 2 seen face-on, with x across and z up. The line $\frac{{x - 5}}{3} = \frac{{z - 1}}{4}$ runs through it, from (5, 2, 1) at t = 0 to (8, 2, 5) at t = 1: 3 across and 4 up for each step, while y stays 2.

Through two points

For the line through A(1, 2, 3) and B(3, 3, 7), use A as the point and AB = (2, 1, 4) as the direction. The Cartesian form is $\frac{{x - 1}}{2} = \frac{{y - 2}}{1} = \frac{{z - 3}}{4}$, the line above.

Check B: $\frac{{3 - 1}}{2} = 1$, $\frac{{3 - 2}}{1} = 1$ and $\frac{{7 - 3}}{4} = 1$. All equal, at t = 1.

Back to vector form

To go the other way, set each fraction equal to a parameter. From $\frac{{x - 2}}{3} = \frac{{y + 1}}{3} = \frac{{z - 3}}{{-2}} = \lambda$: $x = 2 + 3\lambda$, $y = -1 + 3\lambda$ and $z = 3 - 2\lambda$, so $r = \left(2, -1, 3\right) + \lambda\left(3, 3, -2\right)$.

The usual mistakes

Swapping the roles. The point goes on top, subtracted; the direction goes underneath, dividing. $\frac{{x - 2}}{1} = \frac{{y - 1}}{2} = \frac{{z - 4}}{3}$ is not the line r = (1, 2, 3) + t(2, 1, 4).

Writing a plus. Rearranging x = 1 + 2t subtracts 1, so the top is x − 1, not x + 1.

Reading the minus signs into the point. In $\frac{{x - 1}}{2}$, the point’s x-coordinate is 1, not −1.

Reading a line that is not in standard form. In $\frac{{3 - z}}{2}$ the direction’s z-component is −2, not 2.

A zip wire

In the application below, a zip wire is given in Cartesian form. Setting each fraction equal to $\lambda$ turns it back into coordinates, z = 0 finds the anchor on the ground, and the position of a fence post gives the height of the wire above it.