A vector has magnitude and direction, and can be written in i, j, k form or as components. Its magnitude comes from three-dimensional Pythagoras, a unit vector is the vector divided by its magnitude, and the vector from A to B is b − a. A straight line through a point with position vector a, in the direction d, has vector equation r = a + λd.
Key vector results
- Magnitude: |a| = √(a₁² + a₂² + a₃²)
- Unit vector in the direction of a: a / |a|
- Vector from A to B: b − a (end minus start)
- Distance AB = |b − a|
- Line through A in direction d: r = a + λd
The magnitude of a vector
The magnitude of a vector is its length. For a vector with components a₁, a₂ and a₃, it is √(a₁² + a₂² + a₃²), which is Pythagoras' theorem extended to three dimensions. The same rule works in two dimensions with two components, and in either i, j, k or column form.
Unit vectors
A unit vector has magnitude 1. To find the unit vector in the direction of a, divide a by its own magnitude. A unit vector describes direction only. To build a vector of length k in the same direction, multiply the unit vector by k.
Position vectors and the vector between two points
The position vector of a point A is the vector from the origin to A, written a. The vector from A to B is found by subtracting position vectors, end minus start: b − a. The distance between the points is the magnitude of that vector.
The vector equation of a line
A straight line is fixed by one point on it and its direction. If a is the position vector of a point on the line and d is a direction vector, every point on the line has position vector r = a + λd for some value of the parameter λ. The line through two points A and B can use a as the point and b − a as the direction.
Does a point lie on a line?
To test a point P, set its coordinates equal to the components of a + λd. Solve one component equation for λ, then check that the same λ works in the other components. If it does, P lies on the line; if any component fails, it does not.
Three points A, B and C are collinear if the vector from A to B is a multiple of the vector from B to C, because the vectors are parallel and share the point B.
Worked examples
Example 1. Find |a| for a = 2i + 3j − 6k.
- Square and add the components: 4 + 9 + 36 = 49.
- Take the square root.
Answer: |a| = 7.
Marks: M1 for the method, A1 for 7.
Example 2. Find the unit vector in the direction of a = 2i − j + 2k.
- |a| = √(4 + 1 + 4) = 3.
- Divide each component by 3.
Answer: (2/3)i − (1/3)j + (2/3)k.
Marks: M1 for |a| = 3, A1 for the unit vector.
Example 3. A is (1, 4) and B is (5, 2). Find the vector AB and the distance AB.
- Vector AB = b − a = (5 − 1)i + (2 − 4)j = 4i − 2j.
- AB = √(16 + 4) = √20.
Answer: 4i − 2j; AB = 2√5.
Marks: M1 for b − a, A1 for each answer.
Example 4. Write a vector equation of the line through A(2, −1, 3) parallel to 4i + j − 2k.
- Point: a = 2i − j + 3k. Direction: d = 4i + j − 2k.
- Use r = a + λd.
Answer: r = (2i − j + 3k) + λ(4i + j − 2k).
Marks: M1 for the form, A1 for the equation.
Example 5. Determine whether P(7, 5, 10) lies on the line through A(1, 2, 1) and B(3, 3, 4).
- Direction: b − a = 2i + j + 3k, so the line is r = (i + 2j + k) + λ(2i + j + 3k).
- x: 1 + 2λ = 7 gives λ = 3.
- y: 2 + 3 = 5, which matches. z: 1 + 9 = 10, which matches.
Answer: Yes, P lies on the line, at λ = 3.
Marks: M1 for λ from one component, M1 for checking the others, A1 for the conclusion.
Common mistakes
- Subtracting the wrong way. The vector from A to B is b − a, end minus start.
- Forgetting to square negative components. (−6)² = 36; every square is positive.
- Checking only one component. A point is on a line only if the same λ works in every component.
- Using a position vector as the direction. For the line through A and B, the direction is b − a, not b.
- Not dividing every component by the magnitude. A unit vector needs each component divided by |a|.
How vector lines are examined
Vectors in three dimensions, including the vector equation of a line, form the last chapter of Pure 4 (WMA14) in Edexcel International A-Level. In single A-Level Maths (9MA0), magnitudes and position vectors are covered, but vector equations of lines are Further Maths content. Typical questions ask you to:
- find a vector between two points and its exact magnitude
- find a unit vector in a given direction
- write a vector equation of a line through two points
- decide whether a given point lies on a line
The Pure 4 vectors exam questions include these skills with full mark schemes. Use the Pure 4 vectors worksheet or the A-Level vectors revision engine for more practice.
Next, see parallel, intersecting and skew lines. For the GCSE introduction, see vectors.
Frequently asked questions
What is the magnitude of a 3D vector?
Square each component, add them and take the square root: |a| = √(a₁² + a₂² + a₃²).
What is the vector equation of a line?
r = a + λd, where a is the position vector of a point on the line, d is a direction vector and λ is a parameter.
How do you find a unit vector?
Divide the vector by its magnitude, so that the result has length 1.