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Additional Mathematics · Lesson

Explain direction in a vector equation

A vector equation of a line has two vectors in it, and it is easy to use the wrong one.

On this page
  1. How do you read a vector equation?
  2. Worked example
  3. The mistake to watch for
  4. Check yourself
  5. Where this leads next

In r = p + λd, the vector p says where the line passes through and d says which way it runs. Any nonzero multiple of d gives the same direction, so two lines are parallel when their direction vectors are multiples of each other.

This lesson completes the toolkit for two-dimensional vector proofs.

How do you read a vector equation?

  1. Point: put λ = 0 and you get p, a point on the line.
  2. Direction: d is the direction. Simplify by dividing by a common factor.
  3. Gradient (for column vectors (x, y)): y ÷ x.
  4. Test a point: solve for λ from one coordinate, then confirm the other.
  5. Parallel test: check whether one direction is a multiple of the other.

Worked example

Line l: r = (2, −1) + λ(4, −6). (a) State a simpler direction vector and the gradient.

(b) Is (8, −10) on l? (c) Is (6, −8) on l?

(a) (4, −6) = 2(2, −3), so a simpler direction is (2, −3). The gradient is −3 ÷ 2 = −3/2.

(b) x: 2 + 4λ = 8, so λ = 3/2. y: −1 − 6λ = −1 − 9 = −10. It matches the y value, so (8, −10) is on l.

(c) x: 2 + 4λ = 6, so λ = 1. y: −1 − 6(1) = −7, but the point has y = −8. The two coordinates disagree, so (6, −8) is not on l.

Line l with the two test pointsGraph of line l through (2, −1) with direction (4, −6). The point (8, −10) lies on l. The point (6, −8) lies below the line, because l passes through (6, −7). 246810−2−4−6−8−10−12xy (2, −1)(6, −7)(8, −10) on l(6, −8) not on l l: r = (2, −1) + λ(4, −6)
Line l: r = (2, −1) + λ(4, −6), drawn to scale. At x = 6 the line has y = −7, so (6, −8) misses it; (8, −10) lies on it.

The mistake to watch for

A common slip is to check only the x value.

Mistaken working: “For (6, −8): 2 + 4λ = 6, so λ = 1. A value of λ exists, so the point is on the line.”

The student never tested y.

A single coordinate can always be solved for some λ. The point is on the line only when the same λ fits both coordinates. Always carry λ across to the second coordinate.

Check yourself

1. Line r = (3, 1) + λ(−2, 5). State a direction vector with positive x part and the gradient.

Show answer

Multiply (−2, 5) by −1 to get (2, −5). Gradient = −5 ÷ 2.

Direction (2, −5), gradient −5/2

2. Are r = (1, 1) + λ(3, −2) and r = (0, 4) + μ(−6, 4) parallel? Are they the same line?

Show answer

(−6, 4) = −2(3, −2), so the lines are parallel. Is (0, 4) on the first? 1 + 3λ = 0 gives λ = −1/3, then y = 1 − 2(−1/3) = 5/3, not 4.

Parallel but not the same line.

3. The line r = (1, 0) + λ(2, k) passes through (7, 6). Find k.

Show answer

x: 1 + 2λ = 7, so λ = 3. y: 0 + 3k = 6.

k = 2

Where this leads next

Put all five skills together in the two-dimensional vector proofs practice set. You can revisit finding an intersection if the parameter work still needs time.

If you can read an equation but hesitate when planning a whole proof, our teachers can help through online one-to-one Additional Mathematics tuition.

Questions people ask

Which part of r = p + λd gives the direction?

The vector d, the one multiplied by λ, gives the direction. The vector p is the position of one point on the line. Changing p slides the line to a new place, while changing d tilts it.

Are (4, −6) and (2, −3) the same direction?

Yes. (4, −6) = 2(2, −3), so both point along the same line. Any nonzero multiple of a direction vector is another direction vector for that line, including negative multiples that point the other way.

How do I test whether a point is on the line?

Set the equation equal to the point and solve for λ from one coordinate. Then check the other coordinate gives the same λ. If it does, the point is on the line. If not, it is not.

Updated:

Your next step

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