Vectors and coordinates

A vector is a directed segment. In physics, velocity, force and displacement are represented by vectors, because for them not only the size (magnitude) matters but also the direction. In geometry, the coordinate method turns a geometric problem into algebraic calculation: without a drawing, using formulas alone, you can find distances and angles and test for parallel and perpendicular lines. This lesson covers the idea of a vector, coordinates, operations, the scalar product and its uses, and the equations of a line and a circle.

§1. The idea of a vector

A vector is a segment with a marked start and end: in AB⃗, A is the start (tail) and B the end (head). A vector is defined by two things: its length (magnitude) |AB⃗| and its direction.

• Equal vectors have the same length and direction; where they are drawn does not matter (a vector may be translated).
• The zero vector 0⃗ has coinciding start and end, length 0 and no defined direction.
• Opposite vectors have equal lengths and opposite directions: BA⃗ = −AB⃗.
• Collinear (parallel) vectors lie on one line or on parallel lines (pointing the same way or opposite ways).

Segment versus vector: the segments AB and BA are the same, but the vectors AB⃗ and BA⃗ are different.

§2. Coordinates, length and midpoint

If A(x₁; y₁) and B(x₂; y₂):

The length formula is Pythagoras’ theorem: the components x and y are the legs and the vector is the hypotenuse. In space a third coordinate is added: |a⃗| = √(x² + y² + z²).
A vector can be written with unit vectors: a⃗ = x·i⃗ + y·j⃗, where i⃗(1; 0) and j⃗(0; 1) are the unit vectors along the axes.

Worked example

A(1; 2), B(4; 6). Find the components and length of AB⃗ and the midpoint of AB.

1. AB⃗ = (4 − 1; 6 − 2) = (3; 4).
2. |AB⃗| = √(9 + 16) = 5.
3. Midpoint: ((1 + 4)/2; (2 + 6)/2) = (2.5; 4).

§3. Operations on vectors

• Addition: add the components. Geometrically — the triangle rule (AB⃗ + BC⃗ = AC⃗: place the tail of the second vector at the head of the first) or the parallelogram rule (the sum of two vectors from a common start is the diagonal of the parallelogram).
• Subtraction: a⃗ − b⃗ = a⃗ + (−b⃗); drawn from a common start, the difference points from the head of b⃗ to the head of a⃗.
• Multiplication by a number: k·a⃗ = (kx; ky); the length is multiplied by |k|, and for k < 0 the direction reverses.

These operations prove geometric facts: for example, the fourth vertex of a parallelogram is D = A + C − B, because AD⃗ = BC⃗.

Worked example

a⃗(2; −4) and b⃗(−3; k) are collinear. Find k.

1. Collinearity condition: 2/(−3) = −4/k.
2. 2k = 12.
3. k = 6.

§4. The scalar (dot) product

The result is a number (not a vector). The two definitions agree because of the cosine rule. The sign of the scalar product tells you the type of angle: positive — acute, zero — right, negative — obtuse. In physics the work done by a force, A = F⃗ · s⃗, is a scalar product.

Worked example

a⃗(3; k) and b⃗(2; −6) are perpendicular. Find k.

1. Perpendicular means a⃗ · b⃗ = 0.
2. 3 · 2 + k · (−6) = 6 − 6k = 0.
3. k = 1.

Worked example

A(1; 1), B(4; 5), C(5; −2). Find angle A of triangle ABC.

1. AB⃗ = (3; 4), AC⃗ = (4; −3).
2. AB⃗ · AC⃗ = 12 − 12 = 0.
3. So ∠A = 90°.

§5. Working with lengths and an angle

If no coordinates are given, only lengths and the angle, square the expression and use the properties of the scalar product:

This looks like a special-product formula and is in fact the cosine rule for the diagonals of a parallelogram.

Worked example

|a⃗| = 5, |b⃗| = 4, and the angle between them is 60°. Find |5a⃗ − b⃗|.

1. Square it: |5a⃗ − b⃗|² = 25|a⃗|² − 10 a⃗·b⃗ + |b⃗|².
2. a⃗·b⃗ = 5 · 4 · ½ = 10.
3. = 25 · 25 − 100 + 16 = 541 → |5a⃗ − b⃗| = √541.

§6. Equations of a line and a circle

The equation of a circle describes the points at distance R from the centre (the distance formula). The vector n⃗(a; b) is perpendicular to the line ax + by + c = 0 (a normal vector). For a circle given in general form, complete the squares to find the centre and radius.

Worked example

Find the centre and radius of the circle x² + y² − 4x + 6y − 3 = 0.

1. Group: (x² − 4x) + (y² + 6y) = 3.
2. Complete the squares: (x − 2)² − 4 + (y + 3)² − 9 = 3 → (x − 2)² + (y + 3)² = 16.
3. Centre (2; −3), R = 4.

Worked example

Find the equation of the line through M(1; 2) perpendicular to y = 2x + 3.

1. Perpendicular condition: 2 · k = −1 → k = −1/2.
2. Substitute M(1; 2) into y = −x/2 + b: 2 = −1/2 + b → b = 5/2.
3. Answer: y = −x/2 + 5/2, i.e. x + 2y − 5 = 0.

Check yourself

1. Find the length of a⃗(−6; 8).

2. Find the scalar product of a⃗(2; 3) and b⃗(4; −1).

3. a⃗(k; 4) ⟂ b⃗(2; −1). Find k.

4. Find the midpoint of the segment with ends (−2; 5) and (4; −1).

5. |a⃗| = 3, |b⃗| = 4, a⃗ ⟂ b⃗. Find |a⃗ + b⃗|.

6. Find the cosine of the angle between a⃗(1; 2) and b⃗(2; 1).

7. In parallelogram ABCD, A(1; 2), B(4; 3), C(6; 7). Find D.

8. |a⃗| = 2, |b⃗| = 3, angle 120°. Find |a⃗ + b⃗|.

Common mistakes

• Finding a vector’s components by subtracting "end from start" the wrong way round.
• Taking |a⃗ + b⃗| to be |a⃗| + |b⃗|.
• Forgetting the 2a⃗·b⃗ term when squaring.
• Pairing the components wrongly in the collinearity condition.
• Writing the result of a scalar product as a vector (it is a number).
• Getting the sign of the centre wrong: in (x + 3)², a = −3.