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

These notes condense Unit P4 section 7 (Vectors, topics 7.1 to 7.7) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification lists P4 as compulsory for IAL Mathematics and Pure Mathematics, with IAS weighting N/A. For full explanations read the study guide, then test yourself with the practice questions.

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Key results

Topic	Result
7.1 Components	a = a₁i + a₂j + a₃k
7.2 Magnitude	|a| = √(a₁² + a₂² + a₃²)
7.2 Unit vector	â = a/|a|
7.3 Scalar multiple	λa is parallel to a; the sign of λ sets the direction
7.4 Between points	AB = OB − OA = b − a
7.5 Distance	d² = (x₁ − x₂)² + (y₁ − y₂)² + (z₁ − z₂)²
7.6 Line, point and direction	r = a + tb
7.6 Line, two points	r = c + t(d − c)
7.7 Scalar product	a·b = a₁b₁ + a₂b₂ + a₃b₃
7.7 Angle	cos ∠AOB = (a·b)/(|a||b|)
7.7 Perpendicular	a·b = 0 for non-zero a, b means perpendicular

Notation to recognise

Symbol	Meaning
a	the vector a
AB	the vector represented by the directed line segment AB
â	unit vector parallel to a
i, j, k	the axis unit vectors
|a| or |AB|	magnitude of a vector
a.b	the scalar product of a and b

In this specification the scalar product is written with a dot. The vector product a × b is a different operation and belongs to Further Pure Mathematics, so do not confuse the two.

Method boxes

Position vector of a point on a segment. If P divides AB so that AP : PB = m : n, then p = a + (m/(m + n))(b − a). The midpoint of AB has position vector (a + b)/2.

Is a point on a line?

1. Equate the i, j and k components with the point.
2. Solve one equation for t.
3. Substitute that t into the other two. All three must hold.

Parallel, intersecting or skew?

1. Compare directions. If one is a scalar multiple of the other, the lines are parallel (identical if one has a point on the other).
2. Otherwise use different parameters t and s and equate components.
3. Solve two equations, test in the third.
4. Third holds: intersect, and substitute to get the point. Third fails: skew.

Angle between two lines.

1. Take the direction vectors d₁ and d₂ only.
2. cos θ = |d₁·d₂| / (|d₁||d₂|) gives the acute angle.
3. If the question asks for the angle between the vectors themselves, drop the modulus on the top and keep the sign.

Foot of the perpendicular from P to a line.

1. Write the general point F on the line in terms of t.
2. Form PF = F − P.
3. Solve PF·d = 0 for t.
4. Distance from P to the line is |PF|.

Small worked reminders

Magnitude and unit vector. For v = i − 2j + 2k, |v| = √(1 + 4 + 4) = 3, so v̂ = (i − 2j + 2k)/3.

Angle between two vectors. With d₁ = i − 2j + 2k and d₂ = 2i + 3j + 6k:

d₁·d₂ = 2 − 6 + 12 = 8
|d₁| = 3,  |d₂| = 7
cos θ = 8/21,  θ = 67.6° (1 d.p.)

Line through two points. For C(−3, 1, 4) and D(1, 4, 2): d − c = 4i + 3j − 2k, so r = (−3i + j + 4k) + t(4i + 3j − 2k).

Intersecting or skew: a second example

Let l₁: r = (3j + k) + t(i − j + 2k) and l₂: r = 6k + s(2i + j − k). The directions are not multiples of each other.

i:  t = 2s
j:  3 − t = s
k:  1 + 2t = 6 − s

From the first two, 3 − 2s = s, so s = 1 and t = 2. In the k equation the left side is 1 + 4 = 5 and the right side is 6 − 1 = 5. The lines intersect at (2, 1, 5). If l₂ started at 7k instead, the right side would be 6 and the lines would be skew.

For the angle, d₁·d₂ = 2 − 1 − 2 = −1 and |d₁| = |d₂| = √6, so cos θ = −1/6 and θ = 99.6°. The acute angle between the lines is 180° − 99.6° = 80.4°.

Collinear points

Three points are collinear when AB is a scalar multiple of BC. The two vectors are parallel and share the point B, so all three points sit on one line. For A(1, 2, 3), B(3, 5, 2) and C(7, 11, 0):

AB = 2i + 3j − k
BC = 4i + 6j − 2k = 2AB

So A, B and C lie on one straight line. State the multiple and the shared point in your conclusion.

Parallelograms and rhombuses

• In a parallelogram ABCD, AB = DC and AD = BC. Use whichever pair gives you the missing vertex quickly.
• A rhombus is a parallelogram with two adjacent sides of equal length, so test |AB| = |BC|.
• The diagonals of a rhombus are perpendicular, so you can test a·b = 0 for the two diagonal vectors.
• The diagonals of any parallelogram bisect each other, so both have the same midpoint.

Writing answers

• Say “end minus start” in your head before you write each vector.
• Show the equations for each component when solving for an intersection, not just the answers.
• Write the conclusion in words: “the lines intersect at …”, “the lines are skew because …”, “the vectors are perpendicular because a·b = 0”.
• Give exact values such as 2√14 when the question asks for them, and decimals to the accuracy requested.

Must-know distinctions

• Position vector or displacement. OA is fixed to the origin. AB is the move from A to B, found as b − a.
• Direction vector or position vector in a line. In r = a + tb, a locates the line and b gives its direction. Only b is used for the angle between lines.
• Parallel or skew. Parallel lines have proportional directions. Skew lines do not, and still never meet.
• Scalar product or magnitude. a·a = |a|², which is a quick way to get a magnitude. The scalar product itself is a number.
• Vector or unit vector. A vector of magnitude 1 only. Divide by |a|.
• Zero scalar product. a·b = 0 means perpendicular only when both vectors are non-zero.

Quick self-test

1. Find |8i + j − 4k|.
2. Write down the unit vector in the direction of 2i − 2j + k.
3. A is (4, −1, 2) and B is (0, 4, 2). Find AB and the distance AB.
4. Show that (1, 5, −1) lies on r = (i + j + k) + t(2j − k).
5. Are r = (2i + k) + t(i − 2j + 3k) and r = (j + 5k) + s(−2i + 4j − 6k) parallel?
6. Find a·b when a = 3i − j + 2k and b = i + 4j − 5k.
7. Find the angle between the vectors i + 2j + 2k and 2i + 3j + 6k, to 1 d.p.
8. Find p so that pi + 3j − 2k is perpendicular to 4i − 2j + pk.
9. Do r = (2i + j) + t(i + j + k) and r = (2i + 3j + 3k) + s(2i − k) intersect? If so, find the point.
10. Find the midpoint of A(−2, 5, 1) and B(6, 1, −3).

Answers

1. √(64 + 1 + 16) = √81 = 9.
2. (2i − 2j + k)/3, since the magnitude is 3.
3. AB = −4i + 5j + 0k, and |AB| = √41.
4. At t = 2 the point is (1, 1 + 4, 1 − 2) = (1, 5, −1). It lies on the line.
5. The second direction is −2 times the first, so yes, parallel. They are different lines: the point (2, 0, 1) would need s = −1 from the i-component of the second line, which gives a j-component of −3, not 0.
6. 3 − 4 − 10 = −11.
7. a·b = 2 + 6 + 12 = 20, |a| = 3, |b| = 7, cos θ = 20/21, so θ = 17.8°.
8. (p, 3, −2)·(4, −2, p) = 4p − 6 − 2p = 2p − 6 = 0, so p = 3.
9. i: 2 + t = 2 + 2s; j: 1 + t = 3; so t = 2, s = 1. k: 2 on the left and 3 − 1 = 2 on the right. Yes, at (4, 3, 2).
10. (2, 3, −1).

Where marks are usually lost

• Writing AB as a − b, which is really BA. Say “end minus start” every time.
• Stopping after two component equations and announcing an intersection.
• Using the same parameter letter in both lines, so that a valid intersection is missed.
• Rounding in the middle of a calculation, so the final decimal is slightly off. Keep surds or unrounded values until the last line.
• Using position vectors rather than direction vectors in the angle between lines.
• Quoting an obtuse angle when the acute angle between lines is wanted, or the reverse when the angle between the vectors is wanted.
• Not showing the scalar product working when asked to show perpendicularity: state a·b = 0 and the arithmetic.
• Calling lines skew without first ruling out parallel.
• Dropping a minus sign in a component when a coordinate is negative.

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).