Vectors, lines and planes
HL vectors in full: column and i, j, k notation, magnitude and unit vectors, the scalar product and angles, vector equations of lines, intersecting, parallel and skew lines, the vector product, equations of planes, and intersections and angles between lines and planes.
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In 30 seconds
- A vector has size and direction. AB = b − a (end minus start), and |v| = √(x2 + y2 + z2).
- Scalar (dot) product: v · w = v1w1 + v2w2 + v3w3 = |v||w| cos θ. It is 0 exactly when the vectors are perpendicular.
- Line: r = a + λb (point a, direction b). In 3D two lines are parallel, intersecting or skew.
- Vector (cross) product: v × w is perpendicular to both v and w, and |v × w| = |v||w| sin θ is the area of the parallelogram.
- Plane: r · n = a · n, or ax + by + cz = d, where n = (a, b, c) is the normal vector.
- This whole note is HL only (AHL 3.12–3.18) and unchanged for 2029.
By the end you can
- Use vector notation, find magnitudes and unit vectors, and work with position vectors. HL
- Use the scalar product to find angles and test for perpendicular vectors. HL
- Write vector, parametric and Cartesian equations of lines, and decide whether two lines are parallel, intersecting or skew. HL
- Find and use the vector product, including for areas. HL
- Write equations of planes in vector and Cartesian form. HL
- Find intersections of lines and planes and of two or three planes, and the angles between them. HL
1Vector basics
- Unit vector in the direction of v: v/|v|. Parallel vectors are multiples of each other: w = kv.
- Notation: (3, 2, 4) as a column vector is the same as 3i + 2j + 4k. Write vectors with an underline or arrow in your working.
A(1, 2, −1) and B(4, −2, 11). Find AB and a unit vector in the direction of AB.
- AB = b − a = (3, −4, 12) A1
- |AB| = √(9 + 16 + 144) = 13 A1
- unit vector 113(3, −4, 12) A1
- Dividing a line: the point P that divides AB in the ratio 1 : 2 has p = a + ⅓AB.
2The scalar product
Find the angle between a = (2, −1, 3) and b = (1, 4, −2).
- a · b = 2 − 4 − 6 = −8 A1
- cos θ = −8√14 √21 M1
- θ = 118° A1
Find p if (p, 2, −1) and (3, p, 4) are perpendicular.
- 3p + 2p − 4 = 0 M1
- p = 0.8 A1
3Equations of lines
- Parametric form: x = a1 + λb1, y = a2 + λb2, z = a3 + λb3. Cartesian form: x − a1b1 = y − a2b2 = z − a3b3.
- Kinematics: r = r0 + tv describes a particle moving with constant velocity v; speed = |v|. Two particles collide only if they are at the same point at the same time.
- The test: if the directions are multiples, the lines are parallel (or the same line). If not, equate the components: if one λ and μ satisfy all three equations they intersect; if not, they are skew.
L1: r = (1, 2, 3) + λ(2, −1, 1) and L2: r = (2, −3, 5) + μ(1, 1, 0). Show that they intersect and find the point.
- x: 1 + 2λ = 2 + μ; y: 2 − λ = −3 + μ M1
- subtract: 3λ − 1 = 5 ⇒ λ = 2, μ = 3 A1
- z: 3 + 2 = 5 ✓, so they meet at (5, 0, 5) R1A1
Find the acute angle between L1 and L2.
- cos θ = ǀ2 − 1 + 0ǀ√6 √2 = 1√12 M1
- θ = 73.2° A1
4The vector product
Find (2, −1, 3) × (1, 4, −2).
- x: (−1)(−2) − (3)(4) = −10 M1
- y: (3)(1) − (2)(−2) = 7; z: (2)(4) − (−1)(1) = 9 A1
- (−10, 7, 9) A1
- Check: dot your answer with both original vectors. (−10, 7, 9) · (2, −1, 3) = −20 − 7 + 27 = 0 ✓.
- Order matters: w × v = −(v × w).
5Equations of planes
Find the Cartesian equation of the plane through A(1, 0, 2), B(2, 1, 0) and C(0, 3, 1).
- AB = (1, 1, −2), AC = (−1, 3, −1) A1
- n = AB × AC = (5, 3, 4) M1A1
- d = n · a = 5 + 0 + 8 = 13, so 5x + 3y + 4z = 13 A1
Find the area of triangle ABC.
- area = ½|AB × AC| = ½√(25 + 9 + 16) M1
- = 5√22 ≈ 3.54 A1
6Intersections and angles
- Line meets plane: substitute the parametric form of the line into the plane's equation and solve for λ.
Find where r = (1, 1, 1) + λ(1, 0, 2) meets the plane x + 2y − z = 6.
- (1 + λ) + 2 − (1 + 2λ) = 6 M1
- λ = −4 A1
- (−3, 1, −7) A1
Find the acute angle between that line and the plane.
- the angle with the normal: d · n = 1 + 0 − 2 = −1 M1
- sin θ = ǀ−1ǀ√5 √6 = 1√30 M1
- θ = 10.5° A1
- Two planes: their line of intersection has direction n1 × n2. Find one point on it by setting one coordinate (for example z = 0) and solving the other two equations.
Find the line of intersection of x + y + z = 6 and x − y + 2z = 5.
- direction = (1, 1, 1) × (1, −1, 2) = (3, −1, −2) M1A1
- z = 0: x + y = 6, x − y = 5 ⇒ (5.5, 0.5, 0) A1
- r = (5.5, 0.5, 0) + λ(3, −1, −2) A1
- Angle between two planes = the angle between their normals (take the acute one). Three planes can meet at a point, along a line (a sheaf), or not at all; solve the 3 × 3 system with the GDC and interpret.
Test yourself — cover the answers first
Quick check
- HL Find |(2, −3, 6)|.
- HL Find (1, 2, 3) · (4, −1, 0).
- HL What does a zero dot product tell you?
- HL Give a normal vector to the plane 2x − y + 5z = 7.
- HL What are skew lines?
- HL Is (1, 0, 0) × (0, 1, 0) equal to k or −k?
Exam-style
- HL Show that L1: r = (1, 0, 0) + λ(1, 1, 0) and L2: r = (0, 1, 3) + μ(0, 1, 1) are skew.
- HL Find the acute angle between the planes 5x + 3y + 4z = 13 and x + y + z = 6.
Show answers
- 7
- 2
- the vectors are perpendicular
- (2, −1, 5)
- lines that are not parallel and do not intersect
- k
- not parallel: (1, 1, 0) is not a multiple of (0, 1, 1)
x: 1 + λ = 0 ⇒ λ = −1; y: −1 = 1 + μ ⇒ μ = −2
z: 0 = 3 + (−2) = 1 ✗, no common point, so skew AG - normals (5, 3, 4) and (1, 1, 1): dot product 12
cos θ = 12√50 √3
θ = 11.5°
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