amazon.co.ukVectors in 2D and 3D: components, magnitude, unit vectors, position vectors, the scalar product, vector equations of lines and geometric applications - A-Level pure maths vectors (Edexcel 9MA0 Topic 10, AQA 7357 Section J) revised with worked examples and mark-scheme logic.
A vector carries magnitude and direction; a scalar carries magnitude only. In component form, a vector in the plane is a = x i + y j (or the column (x, y)), and in three dimensions a = x i + y j + z k (Edexcel 9MA0 Topic 10; AQA 7357 Section J; OCR A H640 14.01; WJEC/Eduqas and CCEA include 3D vectors at A-level). The magnitude is the length: |a| = sqrt(x^2 + y^2) in 2D, extending to sqrt(x^2 + y^2 + z^2) in 3D by three-dimensional Pythagoras. The unit vector in the direction of a is a/|a| - dividing by the magnitude keeps the direction but gives length 1. Questions ask for a unit vector in a given direction, or for the direction angle: in 2D, tan(theta) = y/x with the quadrant read from the signs of the components.
Two vectors are equal exactly when all corresponding components match - the algebraic statement behind geometric arguments. The standard unit vectors i, j, k point along the axes, so (3, 2, 4) = 3i + 2j + 4k; converting between the two notations is routine but worth doing carefully, since a dropped component is a silent error.
Find the magnitude of a = 2i - j + 2k, and a unit vector in the direction of a.
Line 1: |a| = sqrt(2^2 + (-1)^2 + 2^2) = sqrt(4 + 1 + 4) = sqrt(9) = 3 (M1 A1).
Line 2: Unit vector: a/|a| = (2i - j + 2k)/3 = (2/3)i - (1/3)j + (2/3)k (A1).
Line 3: Check: the magnitude of the unit vector is sqrt(4/9 + 1/9 + 4/9) = 1 (A1: verified).
Vector addition is component-wise, with the triangle law as its picture: a + b takes you along a then along b. Subtraction reverses the second vector; scalar multiplication scales the length (and reverses the direction when the scalar is negative). Two results do the geometric work: the displacement from A to B is AB = b - a (end position minus start position), and a point M is the midpoint of AB when its position vector is m = (a + b)/2.
Position vectors turn geometry into algebra. Three points are collinear when the displacement between one pair is a scalar multiple of the displacement between another pair, AND the lines share a point - AB = t BC for some scalar t. Parallelogram problems use the property that diagonals bisect each other, or that opposite sides are equal as vectors: ABCD is a parallelogram when AB = DC.
The points P, Q and R have position vectors p = (1, 2), q = (3, 4) and r = (5, 6). Show that P, Q and R are collinear.
Line 1: PQ = q - p = (3 - 1, 4 - 2) = (2, 2) (M1).
Line 2: QR = r - q = (5 - 3, 6 - 4) = (2, 2) (M1).
Line 3: QR = 1 x PQ, so the displacements are parallel (A1).
Line 4: The displacements share the point Q, so P, Q and R lie on one straight line (A1: collinearity concluded).
Watch the direction discipline: AB means from A to B, so it is b - a, not a - b - reversing it loses the direction and, in angle questions, the sign of the answer. In mechanics, the same arithmetic drives displacement and velocity vectors, so this section is shared ground with the applied papers.
The scalar (dot) product (Edexcel 9MA0 Topic 10; AQA 7357 J2-J3; OCR A H640 14.01) has two equivalent forms: a . b = |a||b|cos(theta) and a . b = x1x2 + y1y2 + z1z2 (the sum of the products of corresponding components). Equating them gives the angle between vectors: cos(theta) = (a . b)/(|a||b|). The dot product is a scalar, not a vector - a common confusion in definitions.
Two special cases do the geometric work: vectors are perpendicular exactly when a . b = 0 (cos(90 degrees) = 0), and parallel when their components are in the same ratio (b = t a, giving |cos(theta)| = 1). Perpendicularity questions ask for an unknown component: set the dot product to zero and solve.
Find the angle between a = (1, 2, 2) and b = (2, 1, 0).
Line 1: a . b = (1)(2) + (2)(1) + (2)(0) = 4 (M1).
Line 2: |a| = sqrt(1 + 4 + 4) = 3 and |b| = sqrt(4 + 1) = sqrt(5) (M1).
Line 3: cos(theta) = 4/(3sqrt(5)) = 0.5963 (A1).
Line 4: theta = arccos(0.5963) = 0.932 radians = 53.4 degrees (A1).
Find the value of x for which a = (2, -1, 3) and b = (1, x, -2) are perpendicular.
Line 1: Perpendicular requires a . b = 0 (M1).
Line 2: (2)(1) + (-1)(x) + (3)(-2) = 2 - x - 6 = -x - 4 (A1).
Line 3: -x - 4 = 0, so x = -4 (A1).
The dot product also powers geometric proofs: a right angle in a triangle is proved by showing two side vectors have zero dot product, which is often faster than gradients or Pythagoras, and it works in 3D where gradients do not.
The line through point A with position vector a and direction d has vector equation r = a + t d, where t is a scalar parameter: as t runs over all reals, r runs along the line. Through two points A and B, the direction is d = b - a, so r = a + t(b - a). The Cartesian form converts the components: for d = (l, m, n), the line is (x - x1)/l = (y - y1)/m = (z - z1)/n - each component divided by its direction component. A zero direction component means that coordinate is constant on the line, so the Cartesian fraction for it is omitted.
In 3D, intersecting lines are found by equating components: two equations solve for t and s, and the third equation must be checked - if it fails, the lines are skew (not parallel, never meeting). The third-equation check is the mark-winning step: skipping it, or claiming intersection from two equations alone, loses the conclusion mark.
Line 1: r1 = (1, 2, 2) + t(1, 1, 0). Line 2: r2 = (1, 5, 1) + s(2, -1, 1). Find their intersection or show they are skew.
Line 1: Equate components: 1 + t = 1 + 2s and 2 + t = 5 - s (M1: equations formed).
Line 2: From the first: t = 2s. Substituting into the second: 2 + 2s = 5 - s, so 3s = 3, s = 1 and t = 2 (A1).
Line 3: Check the third equation: z-coordinate of r1 at t = 2 is 2 + 0 = 2; of r2 at s = 1 is 1 + 1 = 2. They agree (A1: third equation verified).
Line 4: The lines intersect at r = (3, 4, 2) (A1: point stated).
Parallel lines have proportional direction vectors; identical lines have proportional directions AND a shared point - check whether one line's point satisfies the other's equation to distinguish parallel from coincident. These line ideas feed directly into Further Mathematics (where the scalar product finds angles between lines and the shortest-distance problems live) and into mechanics, where velocity vectors follow r = r0 + v t.
Vector questions reward clean algebra and correct notation. Typical applications: proving a quadrilateral is a parallelogram (show AB = DC as vectors), proving a right angle (zero dot product), finding a midpoint or a ratio point (m = (a + b)/2, or r = (b a + a b)/(a + b) style weighted averages for ratio divisions), and kinematics links - a velocity vector v means position moves as r(t) = r0 + v t, so speed is |v| and displacement over time t is v t.
Scalar triple product contexts: the product a . (b x c), where b x c is the vector (cross) product, measures the volume of the parallelepiped spanned by a, b and c - and it is zero exactly when the three vectors are coplanar. The cross product itself, with |b x c| = |b||c|sin(theta), gives the area of the parallelogram spanned by two vectors (half of it, a triangle). The cross product and triple product are formally assessed in Further Mathematics, not single A-level Maths - but the contexts appear as motivation, and knowing that |a||b|sin(theta) measures parallelogram area helps with geometry questions that ask for areas from two side vectors.
A(1, 1), B(4, 2) and C(6, 5) are three vertices of parallelogram ABCD, in order. Find D.
Line 1: In parallelogram ABCD, AB = DC as vectors (opposite sides equal and parallel) (M1: property quoted).
Line 2: AB = b - a = (4 - 1, 2 - 1) = (3, 1) (A1).
Line 3: DC = c - d = (3, 1), so d = c - (3, 1) = (6, 5) - (3, 1) = (3, 4) (A1).
Line 4: Check with the diagonals: midpoint of AC = ((1 + 6)/2, (1 + 5)/2) = (3.5, 3) and midpoint of BD = ((4 + 3)/2, (2 + 4)/2) = (3.5, 3) - equal, confirming D = (3, 4) (A1: verified).
Exam technique: (1) write AB as b - a before doing anything else - direction errors are the biggest single loss; (2) keep surds exact; (3) in 3D line intersections, always check the third equation and state skew or intersecting explicitly; (4) quote the property used (parallelogram, perpendicular, collinear) before the algebra; (5) for angle questions, give the answer in degrees or radians as the question demands, to 1 d.p. or 3 s.f. Board coverage: Edexcel 9MA0 Topic 10, AQA 7357 Section J, OCR A H640 14.01, WJEC/Eduqas AS/A2 vectors, CCEA AS1/A21 - all include 2D and 3D vectors, the scalar product and vector equations of lines.