Vectors in Three Dimensions

Vectors in Three Dimensions

At A-Level, vector work extends from two dimensions into three, covering the scalar (dot) product, equations of lines in 3D, and geometric applications.

Vectors in 3D

A three-dimensional vector is written as a = xi + yj + zk or in column form as (x, y, z). The unit vectors i, j, k point along the positive x, y, z axes respectively.

The magnitude (length) of a = (a₁, a₂, a₃) is:

|a| = √(a₁² + a₂² + a₃²)

A unit vector in the direction of a is a/|a| (the vector divided by its magnitude).

Worked Example: Find the magnitude and unit vector of a = 2i - 3j + 6k.

|a| = √(4 + 9 + 36) = √49 = 7

Unit vector = (2i - 3j + 6k)/7 = (2/7)i - (3/7)j + (6/7)k

Position Vectors and Geometric Results

The position vector of a point P is the vector from the origin O to P, written as OP or p.

Key results:

• The vector from A to B is AB = b - a (position vector of B minus position vector of A)
• The midpoint M of AB has position vector m = (a + b)/2
• A point dividing AB in the ratio m:n has position vector (na + mb)/(m + n)

The Scalar Product (Dot Product)

For two vectors a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃):

a · b = a₁b₁ + a₂b₂ + a₃b₃

This equals the sum of the products of corresponding components.

The geometric definition is:

a · b = |a| |b| cosθ

where θ is the angle between the two vectors (0 ≤ θ ≤ 180°).

Rearranging: cosθ = (a · b) / (|a| |b|)

Key properties:

• If a · b = 0, the vectors are perpendicular (θ = 90°)
• If a · b > 0, the angle between them is acute
• If a · b < 0, the angle between them is obtuse
• a · a = |a|²

Worked Example: Find the angle between a = (1, 2, 3) and b = (4, -1, 2).

a · b = (1)(4) + (2)(-1) + (3)(2) = 4 - 2 + 6 = 8

|a| = √(1 + 4 + 9) = √14, |b| = √(16 + 1 + 4) = √21

cosθ = 8 / (√14 × √21) = 8/√294

θ = arccos(8/√294) ≈ 62.2°

Equation of a Line in 3D

A line through point A (with position vector a) in the direction of vector d has the vector equation:

r = a + td

where t is a scalar parameter.

In component form, if a = (a₁, a₂, a₃) and d = (d₁, d₂, d₃):

x = a₁ + td₁, y = a₂ + td₂, z = a₃ + td₃

Worked Example: Find the equation of the line through A(1, 3, -2) and B(4, 1, 5).

Direction vector AB = b - a = (3, -2, 7)

Line: r = (1, 3, -2) + t(3, -2, 7)

Or: r = (1 + 3t)i + (3 - 2t)j + (-2 + 7t)k

Intersection and Skew Lines

Two lines in 3D can be:

• Parallel (direction vectors are multiples of each other)
• Intersecting (a unique point satisfies both equations)
• Skew (not parallel and do not intersect — only possible in 3D)

To test for intersection, set the two vector equations equal with different parameters (t and s), solve the resulting system, and check all three component equations are consistent.

Worked Example: Do r = (1, 2, 3) + t(1, -1, 2) and r = (3, 0, 5) + s(2, 1, -1) intersect?

From x: 1 + t = 3 + 2s → t - 2s = 2 ... (i)

From y: 2 - t = 0 + s → -t - s = -2 → t + s = 2 ... (ii)

From (i) and (ii): t - 2s = 2 and t + s = 2. Subtracting: -3s = 0, so s = 0, t = 2.

Check z: 3 + 2(2) = 7 and 5 + (-1)(0) = 5. Since 7 ≠ 5, the lines are skew.

Angle Between Two Lines

The angle between two lines is the angle between their direction vectors:

cosθ = |d₁ · d₂| / (|d₁| |d₂|)

The modulus signs ensure we get the acute angle between the lines.

Exam Tips

• When finding AB, always do B minus A (not A minus B).
• To show two vectors are perpendicular, show their scalar product is zero.
• When checking if lines intersect, use different parameter letters (t and s) — using the same letter assumes corresponding points.
• Always verify the third equation when testing for intersection — two equations with two unknowns always have a solution, but the third may be inconsistent (giving skew lines).
• State the direction of the line clearly. Any non-zero scalar multiple of d gives the same line.