The Vector Equation of a Line

A point and a direction

A straight line is fixed by one point on it and the direction it runs in. Give the point A by its position vector, a = (1, 2), and the direction by a vector d = (2, 1), called the direction vector.

Start at A and walk one copy of d: 2 across and 1 up, to (3, 3). Walk a second copy, to (5, 4). Every copy of d points the same way, so every stride lands on the same straight line through A. Walking backward, with −d, lands on it too: (1 − 2, 2 − 1) = (−1, 1).

a = (1, 2) runs from the origin to the start. Two strides of d = (2, 1) reach (3, 3) and then (5, 4), all on one line.

r = a + t d

Every point of the line is reached by going to A and then some number of strides of d. Call that number t. The position vector r of the point is r = a + t d, the vector equation of the line. The number t is called the parameter.

For this line, r = (1, 2) + t(2, 1). At t = 0, r = (1, 2), the point A itself. At t = 1, r = (1 + 2, 2 + 1) = (3, 3), one stride out. At t = 2, r = (1 + 2 × 2, 2 + 2 × 1) = (5, 4), two strides out.

t does not have to be a positive whole number. At t = −1, r = (−1, 1), one stride back behind A. At $t=\frac{1}{2}$t = 1/2, r = (1 + 1, 2 + 0.5) = (2, 2.5), halfway through the first stride. Each value of t gives one point of the line, and every point of the line comes from one value of t.

Here a = (1, 1) and d = (2, 1). At $\lambda =-1$λ = −1, the point is (1, 1) − (2, 1) = (−1, 0), one stride behind A. Drag it along the line: at $\lambda =2$λ = 2 the point is (1 + 4, 1 + 2) = (5, 3), and sweeping $\lambda$λ through every value traces the whole line.

The equation in components

Written one component at a time, r = (1, 2) + t(2, 1) says x = 1 + 2t and y = 2 + t.

The parameter can be removed. From the second equation t = y − 2, and putting that into the first gives x = 1 + 2(y − 2) = 2y − 3, which rearranges to $y=\frac{x}{2}+\frac{3}{2}$y = x/2 + 3/2. The gradient $\frac{1}{2}$1/2 is the rise over the run of d, 1 ÷ 2.

Is a point on the line?

A point lies on the line when one value of t gives both of its coordinates. Is (9, 6) on r = (1, 2) + t(2, 1)? The x-coordinate needs 1 + 2t = 9, so t = 4. Then y = 2 + 4 = 6, which matches, so (9, 6) is on the line, at t = 4.

Is (7, 4) on it? The x-coordinate needs 1 + 2t = 7, so t = 3, but then y = 2 + 3 = 5, not 4. No single value of t fits both coordinates, so (7, 4) is not on the line; it is 1 below the point (7, 5).

Many equations for one line

Doubling the direction vector does not change the line. r = (1, 2) + s(4, 2) starts at the same point and strides in the same direction, twice as far each time: s = 1 gives (5, 4), which is t = 2 on the first equation. Any multiple of d, other than 0, gives the same line, including a negative one, which walks it the other way.

The starting point can change too. r = (3, 3) + t(2, 1) starts from another point of the same line. So two equations that look different can describe one line. To check, see whether the two direction vectors are multiples of each other, and whether a point of one line lies on the other.

From A, d = (2, 1) and 2d = (4, 2) both run along the same line; 2d simply reaches twice as far.

A line through two points

For the line through two points A and B, start at a and take the direction from A to B, which is b − a. With A(1, 2) and B(5, 4), d = (5 − 1, 4 − 2) = (4, 2), and the line is r = (1, 2) + t(4, 2).

Then t = 0 gives A and t = 1 gives B. Values of t from 0 to 1 give the points between them, so the segment AB is the part with $0\le t\le 1$0 ≤ t ≤ 1, and its midpoint is at $t=\frac{1}{2}$t = 1/2: (1 + 2, 2 + 1) = (3, 3).

In three dimensions

The equation keeps the same form with three components. The line through (1, 2, 0) with direction (2, −1, 3) is r = (1, 2, 0) + t(2, −1, 3). At t = 2, r = (1 + 4, 2 − 2, 0 + 6) = (5, 0, 6).

Is (7, −1, 9) on it? The x-coordinate needs 1 + 2t = 7, so t = 3. Then y = 2 − 3 = −1 and z = 0 + 9 = 9, both of which match, so the point is on the line at t = 3.

The usual mistakes

Stopping after one stride. For r = (1, 2) + t(2, 1) at t = 2, the point is (5, 4), not (3, 3): t = 2 walks d twice.

Swapping the roles of a and d. (2, 1) + 2(1, 2) = (4, 5) starts at the wrong point and strides the wrong way. The start is a, and t multiplies d.

Taking the start for the direction. a only says where the line begins; every stride follows d.

Checking one coordinate only. A value of t that fits x must also fit y, and z in three dimensions.

Deciding that a doubled direction gives a different line. A multiple of d points the same way, so the line is the same.

A ship on a straight course

In the application below, the parameter is time. The ship starts at the position vector (2, 1) at noon, and each hour adds the direction vector (3, 4), so the length of that vector is the distance covered in an hour: the speed. Testing the buoy and the reef is testing whether a point is on the line, with one value of the time for both components.