OxfordAQA International AS and A-level Mathematics: Scalar product and perpendicular distance (9660) – Revision Notes

Quick recall for the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section P2.10 (second part): the scalar product, the angle between two lines, and the foot of the perpendicular and perpendicular distance from a point to a line. This content belongs to International A-level Unit P2 and is not examined at International AS. For full explanations, use the study guide.

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Key results

Result	Statement
Component form	(x, y, z).(a, b, c) = xa + yb + zc
Geometric form	u.v = |u||v| cos θ, with 0° ≤ θ ≤ 180°
Angle	cos θ = (u.v) / (|u||v|)
Perpendicular	u.v = 0 (both vectors non-zero)
Length	u.u = |u|²
Parallel, same way	u.v = |u||v|
Parallel, opposite way	u.v = −|u||v|
Sign	u.v > 0 acute, u.v < 0 obtuse
Rules	u.v = v.u and u.(v + w) = u.v + u.w
Line	r = a + td; direction d, point a
2D line y = mx + c	direction (1, m)
2D line ax + by = c	direction (b, −a)

The scalar product is a number. The answer is never a vector.

Terms to know

• Scalar product: the number xa + yb + zc found from two vectors. It is also called the dot product.
• Direction vector: any non-zero vector parallel to the line. Multiples of it describe the same direction.
• Foot of the perpendicular: the point on the line nearest to P, where PX meets the line at a right angle.
• Perpendicular distance: the length PX, the shortest distance from P to the line.
• Skew lines: lines in three dimensions that are neither parallel nor intersecting. They still have an angle between them.

Method: angle between two lines

1. Read off the direction vectors d₁ and d₂ and set aside the points a.
2. Find d₁.d₂, |d₁| and |d₂|.
3. cos θ = (d₁.d₂) / (|d₁||d₂|).
4. If the answer is obtuse and the question asks for the angle between the lines, give the acute one, 180° − θ.

Works for skew lines. Works for parallel lines (0°) and perpendicular lines (90°).

Quick reminder: directions (1, 1, 0) and (0, 1, 1). Scalar product 1, magnitudes √2 and √2, so cos θ = 1/2 and θ = 60°.

Short worked reminder (three dimensions)

Direction vectors (6, 2, 3) and (0, −4, 3).

d₁.d₂ = 0 − 8 + 9 = 1
|d₁| = √(36 + 4 + 9) = 7
|d₂| = √(0 + 16 + 9) = 5
cos θ = 1/35, so θ = 88.4° (1 d.p.)

The scalar product is positive, so this is already the acute angle. The lines are almost, but not exactly, perpendicular.

Short worked reminder (an unknown constant)

Lines with directions (1, 2, 2) and (p, 0, 1) meet at 45°. Find p.

d₁.d₂ = p + 2
|d₁| = 3, |d₂| = √(p² + 1)
(p + 2) / (3√(p² + 1)) = 1/√2
2(p + 2)² = 9(p² + 1)
7p² − 8p + 1 = 0
(7p − 1)(p − 1) = 0
p = 1/7 or p = 1

Squaring can introduce roots that give the supplementary angle. Here p + 2 is positive for both values, so both give 45°.

Method: foot of the perpendicular

For P and line l: r = a + td:

1. X = a + td, written as coordinates in t.
2. PX = X − P, in terms of t.
3. Solve PX.d = 0 for t.
4. Substitute t into X. That point is the foot.

The perpendicular distance is |PX| at that t.

Short worked reminder (two dimensions)

Find the foot and the distance from P(3, 3) to the line r = (1, −3) + t(2, 1).

X = (1 + 2t, −3 + t)
PX = (2t − 2, t − 6)
PX.d = 2(2t − 2) + (t − 6) = 5t − 10 = 0, so t = 2
X = (5, −1)
PX = (2, −4), |PX| = √(4 + 16) = √20 = 2√5

The foot is (5, −1) and the distance is 2√5, about 4.47.

Alternative: no foot needed

If A is on the line and d is its direction:

distance² = |AP|² − (AP.d / |d|)²

Use this when the question asks only for the distance. If it asks for the foot, you still need t.

Must-know distinctions

• Scalar product versus vector: one is a number, the other a vector.
• Direction vector versus position vector: only direction vectors give the angle between lines.
• Angle between vectors versus angle between lines: vectors can give an obtuse angle, lines are quoted as acute (or 90°).
• Perpendicular distance versus distance to a point on the line: only the foot gives the shortest distance.
• Perpendicular test versus parallel test: perpendicular needs u.v = 0. Parallel needs one vector to be a multiple of the other.
• Unknown constants: set u.v = 0 for perpendicular. For a given angle, use the cosine form and expect a quadratic after squaring, then check which roots fit the sign of the scalar product.

Exam habits

• Write the vectors in full before the scalar product, with brackets round negatives.
• Keep surds exact until the final line, then give the decimal the question asks for.
• Show cos θ = … as its own line before you take the inverse cosine.
• Test your foot: PX.d should come out as 0.
• Say what you are finding. A line such as “direction vectors are…” makes your method clear.

Triangle checklist

• Angle ABC: vectors BA and BC, both starting at B.
• Right angle at A: AB.AC = 0.
• Area with a right angle at A: ½ × AB × AC.
• Area in general: ½ × BC × AD, where D is the foot of the perpendicular from A to BC.

Short worked reminder (angle in a triangle)

A(2, 0, 1), B(4, 2, 2) and C(3, −1, 3). Find angle BAC.

AB = (2, 2, 1)     AC = (1, −1, 2)
AB.AC = 2 − 2 + 2 = 2
|AB| = 3, |AC| = √6
cos A = 2 / (3√6) = √6/9
A = 74.2° (1 d.p.)

Both vectors start at A, which is the vertex of the angle. Using AB and CA instead would give the supplement.

Quick self-test

1. Evaluate (2, −3, 1).(4, 1, −5). What does your answer tell you?
2. Find |(2, −3, 6)|.
3. |a| = 4, |b| = 5 and the angle between a and b is 60°. Find a.b.
4. Find the angle between (1, 2, 2) and (0, 3, 4), to 1 d.p.
5. Find k so that (3, k, 1) and (k, 2, −3) are perpendicular.
6. Lines have directions (1, 2, −2) and (−1, 0, 1). Find the acute angle between the lines.
7. Find the foot and distance from P(3, 2, 6) to the line r = (0, 1, 2) + t(1, 2, 0).
8. If u.v = −|u||v|, what is the angle between u and v?
9. P is 13 units from A on a line. The part of AP along the line has length 12. How far is P from the line?
10. Why can the angle between two lines never be quoted as 120°?
11. Find the acute angle between y = x + 2 and y = 3x − 1, to 1 d.p.
12. Find the foot of the perpendicular from (4, 0) to the line y = x, and the distance.

Answers

1. 8 − 3 − 5 = 0, so the vectors are perpendicular.
2. √(4 + 9 + 36) = 7.
3. 4 × 5 × cos 60° = 10.
4. Scalar product 14, magnitudes 3 and 5, cos θ = 14/15, so θ = 21.0°.
5. 3k + 2k − 3 = 0, so k = 3/5.
6. Scalar product −1 − 2 = −3, magnitudes 3 and √2, cos θ = −1/√2, so the vectors are at 135° and the lines at 45°.
7. t = 1 gives the foot (1, 3, 2). PX = (2, −1, 4) has scalar product 0 with (1, 2, 0). Distance √21.
8. 180°. The vectors point in opposite directions.
9. √(13² − 12²) = 5.
10. Two lines make a pair of supplementary angles, 60° and 120°, and the convention is to quote the acute one, 60°.
11. Directions (1, 1) and (1, 3). Scalar product 4, magnitudes √2 and √10, cos θ = 4/√20 = 2/√5, so θ = 26.6°.
12. The line is r = t(1, 1). X = (t, t) and PX = (t − 4, t), so PX.d = 2t − 4 = 0 and t = 2. The foot is (2, 2) and the distance is √(4 + 4) = 2√2.

Where marks are usually lost

• Using the position vectors of the lines instead of their direction vectors.
• Forgetting the denominator |u||v| when finding cos θ.
• Sign slips in the scalar product, especially with negative components. Write each product in brackets first.
• Giving the obtuse angle when the question asks for the angle between two lines.
• Solving PX.d = 0 correctly but not substituting back to give the foot as a point.
• Presenting |AP| (a point on the line) as the perpendicular distance.
• Rounding in the middle of a chain of working, then giving a final answer that is slightly off.
• Using the scalar product to test for parallel lines. It tests for perpendicular ones.
• Claiming a result with no working when the question says show that. Write every line.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section P2.10 (Vectors), published by OxfordAQA. For International AS exams from May/June 2018 and International A-level exams from May/June 2019.