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: $t=\frac{\left(x-1\right)}{2}$t = (x − 1)/2. From y = 2 + t: $t=\frac{\left(y-2\right)}{1}$t = (y − 2)/1. From z = 3 + 4t: $t=\frac{\left(z-3\right)}{4}$t = (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: $\frac{\left(x-1\right)}{2}=\frac{\left(y-2\right)}{1}=\frac{\left(z-3\right)}{4}$(x − 1)/2 = (y − 2)/1 = (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): $\frac{\left(5-1\right)}{2}=2$(5 − 1)/2 = 2, $\frac{\left(4-2\right)}{1}=2$(4 − 2)/1 = 2 and $\frac{\left(11-3\right)}{4}=2$(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}$7/4, not 2, so that point is not on the line.

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

Reading it back

In general, the line through the point $\left(a_1,a_2,a_3\right)$(a₁, a₂, a₃) with direction $\left(d_1,d_2,d_3\right)$(d₁, d₂, d₃) is $\frac{\left(x-a_1\right)}{d_1}=\frac{\left(y-a_2\right)}{d_2}=\frac{\left(z-a_3\right)}{d_3}$(x − a₁)/d₁ = (y − a₂)/d₂ = (z − a₃)/d₃. The numbers taken away on top give the point; the numbers underneath give the direction.

Watch the signs. $\frac{\left(x+2\right)}{3}=\frac{\left(y-1\right)}{\left(-2\right)}=\frac{z}{5}$(x + 2)/3 = (y − 1)/(−2) = z/5 is $\frac{\left(x-\left(-2\right)\right)}{3}=\frac{\left(y-1\right)}{\left(-2\right)}=\frac{\left(z-0\right)}{5}$(x − (−2))/3 = (y − 1)/(−2) = (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 $\frac{\left(2x-4\right)}{6}=\frac{\left(y+1\right)}{3}=\frac{\left(3-z\right)}{2}$(2x − 4)/6 = (y + 1)/3 = (3 − z)/2.

The first fraction is $\frac{2\left(x-2\right)}{6}=\frac{\left(x-2\right)}{3}$2(x − 2)/6 = (x − 2)/3. The third is $\frac{\left(3-z\right)}{2}=-\frac{\left(z-3\right)}{2}=\frac{\left(z-3\right)}{\left(-2\right)}$(3 − z)/2 = −(z − 3)/2 = (z − 3)/(−2). So the line is $\frac{\left(x-2\right)}{3}=\frac{\left(y+1\right)}{3}=\frac{\left(z-3\right)}{\left(-2\right)}$(x − 2)/3 = (y + 1)/3 = (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: $\frac{\left(x-5\right)}{3}=\frac{\left(z-1\right)}{4}$(x − 5)/3 = (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 $\frac{\left(x-5\right)}{3}=\frac{\left(z-1\right)}{4}$(x − 5)/3 = (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 $\frac{\left(x-1\right)}{2}=\frac{\left(y-2\right)}{1}=\frac{\left(z-3\right)}{4}$(x − 1)/2 = (y − 2)/1 = (z − 3)/4, the line above.

Check B: $\frac{\left(3-1\right)}{2}=1$(3 − 1)/2 = 1, $\frac{\left(3-2\right)}{1}=1$(3 − 2)/1 = 1 and $\frac{\left(7-3\right)}{4}=1$(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 $\frac{\left(x-2\right)}{3}=\frac{\left(y+1\right)}{3}=\frac{\left(z-3\right)}{\left(-2\right)}=\lambda$(x − 2)/3 = (y + 1)/3 = (z − 3)/(−2) = λ: $x=2+3\lambda$x = 2 + 3λ, $y=-1+3\lambda$y = −1 + 3λ and $z=3-2\lambda$z = 3 − 2λ, so $r=\left(2,-1,3\right)+\lambda \left(3,3,-2\right)$r = (2, −1, 3) + λ(3, 3, −2).

The usual mistakes

Swapping the roles. The point goes on top, subtracted; the direction goes underneath, dividing. $\frac{\left(x-2\right)}{1}=\frac{\left(y-1\right)}{2}=\frac{\left(z-4\right)}{3}$(x − 2)/1 = (y − 1)/2 = (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 $\frac{\left(x-1\right)}{2}$(x − 1)/2, the point’s x-coordinate is 1, not −1.

Reading a line that is not in standard form. In $\frac{\left(3-z\right)}{2}$(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 $\lambda$λ 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.