Pearson Edexcel International Advanced Level Mathematics: Vectors, lines and the scalar product (YMA01)

This guide teaches Unit P4 section 7 (topics 7.1 to 7.7, Vectors) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification states that P4 is a compulsory unit for IAL Mathematics and Pure Mathematics, and that its IAS weighting is N/A, so this content is not part of IAS. Coordinate geometry from Pure Mathematics 1 and Pure Mathematics 2 is useful background, since lines in the (x, y) plane are the two-dimensional case of this work.

Use it with the revision notes and the practice questions. The course hub is /boards/edexcel/a-level/mathematics/, the printable checklist is /checklists/edexcel/a-level/mathematics/, and the free 10-minute diagnostics show where to start.

What section 7 covers

Ref	You must be able to
7.1	Use vectors in two and three dimensions, written with i, j and k
7.2	Find the magnitude |a| and a unit vector in the direction of a
7.3	Add vectors and multiply by scalars, and explain both geometrically
7.4	Use position vectors, with OB − OA = AB = b − a
7.5	Find the distance between two points
7.6	Write vector equations of lines, r = a + tb and r = c + t(d − c), and decide whether two lines are parallel, intersecting or skew
7.7	Use the scalar product to find the angle between two lines and to test for perpendicular vectors

Vectors, magnitude and unit vectors (7.1 to 7.3)

A vector has magnitude and direction. In three dimensions, the vectors i, j and k have length 1 and point in the positive x, y and z directions, so a = a₁i + a₂j + a₃k. Two vectors are equal when all three components match.

• Add and subtract component by component.
• Multiplying by a scalar λ multiplies every component. The result is parallel to the original, in the same direction if λ > 0 and the opposite direction if λ < 0.
• Geometrically, a + b is found by joining the vectors head to tail, and a − b is a + (−b).

The magnitude is |a| = √(a₁² + a₂² + a₃²). A vector with magnitude 1 is a unit vector. To get one parallel to a, divide a by its own length: â = a/|a|. The notation |AB| means the magnitude of the vector AB.

Worked example 1. Let a = 12i + 4j − 3k.

|a| = √(144 + 16 + 9) = √169 = 13
â = (12i + 4j − 3k)/13

A vector of magnitude 26 in the opposite direction to a is −2a, because |−2a| = 2 × 13 = 26. So it is −24i − 8j + 6k.

Position vectors and distance (7.4 and 7.5)

The position vector of a point A is OA = a, the vector from the origin O to A. For points A and B,

AB = OB − OA = b − a

Read this as “end minus start”. The vector BA is the negative of AB.

The distance between (x₁, y₁, z₁) and (x₂, y₂, z₂) is d, where

d² = (x₁ − x₂)² + (y₁ − y₂)² + (z₁ − z₂)²

This is |AB|, the magnitude of the vector between the points. It does not matter which way round you subtract, because the components are squared.

Worked example 2. A is (3, −2, 4) and B is (9, 2, 16).

AB = (9 − 3)i + (2 − (−2))j + (16 − 4)k = 6i + 4j + 12k
|AB|² = 36 + 16 + 144 = 196, so |AB| = 14

The midpoint M has position vector (a + b)/2 = (6, 0, 10). The point P that splits AB in the ratio 3 : 1, measured from A, lies three quarters of the way along:

p = a + (3/4)AB = (3i − 2j + 4k) + (3/4)(6i + 4j + 12k) = 7.5i + j + 13k

Worked example 3. C is (−1, 2, 0) and D is (3, −2, 5). Then CD = 4i − 4j + 5k, and |CD| = √(16 + 16 + 25) = √57. Leave a surd like this unless the question asks for a decimal.

Vector equations of lines (7.6)

A line is fixed by one point on it and a direction. If A has position vector a and the line is parallel to b, then

r = a + tb

where t is a scalar parameter. Each value of t gives the position vector of one point on the line. The vector b is the direction vector. Any non-zero multiple of b describes the same line direction, so 2b and −b are equally valid.

If you know two points C and D, the direction is d − c, and the line is

r = c + t(d − c)

Here t = 0 gives C and t = 1 gives D. For C and D in worked example 3, the line is r = (−i + 2j) + t(4i − 4j + 5k).

To test whether a point lies on a line, equate the three components. You need a single value of t that works for all three.

Parallel, intersecting or skew

Two lines in three dimensions fall into exactly one of these cases:

1. Parallel: the direction vectors are scalar multiples of each other.
2. Intersecting: not parallel, and there is a point on both lines.
3. Skew: not parallel, and no common point.

In two dimensions, non-parallel lines always meet. In three dimensions they can miss each other, which is why skew lines exist.

Method for intersection. Write the lines with different parameters, say t and s. Equate the i, j and k components. Solve two of the three equations for t and s, then substitute into the third as a check. If it holds, the lines meet, and you substitute back to get the point. If it fails, the lines are skew (given they are not parallel).

Worked example 4. Let

l₁: r = (i + 2j − k) + t(3i − j + 2k)
l₂: r = (2i − 5j + 5k) + s(i + 3j − 2k)

The directions 3i − j + 2k and i + 3j − 2k are not multiples of each other, so the lines are not parallel. Equate components:

i:  1 + 3t = 2 + s
j:  2 − t  = −5 + 3s
k: −1 + 2t = 5 − 2s

The i equation gives s = 3t − 1. Put that in the j equation: 2 − t = −5 + 9t − 3, so 10 = 10t, t = 1 and s = 2. Check k: left side −1 + 2 = 1, right side 5 − 4 = 1. It holds, so the lines intersect. The point is (1 + 3, 2 − 1, −1 + 2) = (4, 1, 1).

Now change l₂ to start at 2i − 5j + 6k, keeping the same direction. The i and j equations still give t = 1, s = 2. The k equation now reads 1 on the left and 6 − 4 = 2 on the right. It fails, so the new line and l₁ are skew.

Parallel lines can also be the same line. Two parallel lines are identical only if a point of one lies on the other, so check that as well when a question asks whether they coincide.

The scalar product (7.7)

For a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, the scalar product (written a.b in the specification) is

a·b = a₁b₁ + a₂b₂ + a₃b₃

The result is a number, not a vector. For OA = a and OB = b, the specification gives

cos ∠AOB = (a·b)/(|a||b|)

The same formula gives the angle θ between any two vectors placed tail to tail, and can be rearranged as a·b = |a||b| cos θ.

If a and b are non-zero and a·b = 0, then cos θ = 0 and the vectors are perpendicular. This is the standard test, and it also works in reverse.

Worked example 5 (angle in a triangle). B is (2, 1, 3), A is (5, 3, 9) and C is (4, −2, 9). Find angle ABC.

BA = 3i + 2j + 6k,  |BA| = √(9 + 4 + 36) = 7
BC = 2i − 3j + 6k,  |BC| = √(4 + 9 + 36) = 7
BA·BC = 6 − 6 + 36 = 36
cos ∠ABC = 36/(7 × 7) = 36/49
∠ABC = 42.7° (1 d.p.)

Both vectors must start at B, because the angle is at B. Using AB instead of BA would change the sign of one vector and give the supplementary angle.

Angle between two lines

Use the direction vectors only. The position vectors a and c in the equations play no part in the angle.

Worked example 6. For l₁ and l₂ in worked example 4, d₁ = 3i − j + 2k and d₂ = i + 3j − 2k.

d₁·d₂ = 3 − 3 − 4 = −4
|d₁| = √14,  |d₂| = √14
cos θ = −4/14 = −2/7,  θ = 106.6° (1 d.p.)

The directions make 106.6°, so the lines make the acute angle 180° − 106.6° = 73.4° with each other. When a question asks for the angle between two lines, give the acute angle unless the context says otherwise.

Perpendicular from a point to a line

To find the foot F of the perpendicular from P to a line with direction d, write F in terms of t, form PF, and solve PF·d = 0.

Worked example 7. P is (8, 3, 3). Find the foot of the perpendicular on l₁ from worked example 4, and the distance from P to l₁.

F = (1 + 3t, 2 − t, −1 + 2t)
PF = F − P = (3t − 7)i − (t + 1)j + (2t − 4)k
PF·d₁ = 3(3t − 7) + (t + 1) + 2(2t − 4) = 14t − 28 = 0, so t = 2
F = (7, 0, 3)
PF = −i − 3j + 0k, so the distance is |PF| = √10

Common errors

• Subtracting the wrong way round, so AB becomes BA. Write “end minus start” before you calculate.
• Calling a vector a unit vector without dividing every component by the magnitude.
• Using one parameter for both lines. Use t and s, or you will force the two points to share a parameter value.
• Declaring lines intersecting after solving only two equations. The third equation is the test.
• Forgetting that skew lines must also be non-parallel.
• Giving an obtuse angle between lines when the question wants the angle between the lines themselves.
• Taking the angle at the wrong vertex by using vectors that do not both start there.

Next steps

Try the practice questions, then keep the revision notes for the final weeks.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. Unit P4, section 7 (Vectors).