Loav · VCE Specialist Mathematics · Units 3 & 4
S34-V02 · Vectors · Products and geometry
1. The dot product measures alignment
Two vectors can have the same lengths yet interact differently because of the angle between them. The dot product combines magnitude and alignment into a scalar. For vectors a and b with components (a₁, a₂, a₃) and (b₁, b₂, b₃), multiply corresponding components and add. There are no cross terms.
a · b = |a||b| cos θ, for nonzero vectors and 0 ≤ θ ≤ π.
The geometric formula explains the sign. A positive dot product corresponds to an acute angle, zero to a right angle, and a negative result to an obtuse angle. The angle is measured between the vectors placed tail to tail. Translating an arrow does not change that angle, but reversing one arrow replaces θ by π − θ and reverses the sign of the dot product.
The component formula can also be deduced from the geometric definition. Each coordinate unit vector has dot product one with itself, because its length is one and its self-angle is zero. Distinct coordinate unit vectors have dot product zero because they are perpendicular. Expand (a₁i + a₂j + a₃k) · (b₁i + b₂j + b₃k) using distributivity. Every mixed-axis term vanishes; the three matching-axis terms remain as a₁b₁ + a₂b₂ + a₃b₃. This explains why the formula depends on perpendicular unit coordinate directions.
The component formula remains valid when a vector is zero. However, an angle involving the zero vector is undefined. Therefore the statement “dot product zero means perpendicular” needs nonzero vectors when it describes two directions. The condition alone is not enough to assign a direction to a zero displacement.
The product is commutative and distributes over addition. In particular, a · a = |a|2. Expanding (a + b) · (a + b) gives |a|2 + 2a · b + |b|2. This is why the magnitude of a sum is generally not the sum of magnitudes: its square contains an angle-dependent term.
2. Angles and perpendicularity need different decisions
To find an angle between nonzero vectors, divide the dot product by the product of magnitudes and then apply inverse cosine. Preserve the exact fraction or surd until the final numerical evaluation. State whether the answer is in degrees or radians. In exact arithmetic the cosine input lies between −1 and 1; a value outside this interval signals a calculation error or a tiny floating-point error that needs investigation.
To test perpendicularity, simply show that the dot product is zero. To test parallelism, show that one vector is a scalar multiple of the other. A dot product of one does not generally mean parallel: the lengths may differ. Only for unit vectors does a dot product of one indicate identical direction, and a dot product of negative one indicate opposite direction.
There is also a distinction between an angle between directed vectors and the smaller angle between undirected lines. Reversing a line's direction vector leaves the line unchanged, so the smaller line angle uses |a · b| in the numerator. Use the absolute value only when that geometric question calls for it; applying it automatically loses valid obtuse vector angles.
Worked example 1: Find an angle and a projection
Let a = (2, −1, 2) and b = (1, 2, 2). Find their angle and the part of a parallel to b.
The dot product is 2 − 2 + 4 = 4. Both magnitudes are 3, so cos θ = 4/9 and θ = cos−1(4/9), approximately 63.6°. The positive dot product correctly predicts an acute angle.
The vector projection of a onto b is [4/9]b = (4/9, 8/9, 8/9). Subtract it from a to obtain the perpendicular remainder (14/9, −17/9, 10/9). Its dot product with b is (14 − 34 + 20)/9 = 0. Thus the decomposition has both required properties: the first part is parallel to b, and the remainder is perpendicular.
3. Scalar and vector resolutes answer different questions
The signed scalar resolute of a in the direction of nonzero b tells how much of a points along that chosen direction. Take the dot product with the unit direction b/|b|. The answer is a scalar with the same measurement units as a. It can be negative: that means the projected arrow points opposite to b.
Vector resolute or projection: projba = [(a · b)/|b|2]b
The vector resolute multiplies the signed scalar resolute by the unit direction. Its magnitude is the absolute value of the scalar resolute. These formulas therefore have different denominators and different output types. Labelling a bare number as a vector, or attaching a direction vector to the scalar formula without another normalisation, changes the answer.
The perpendicular component is a − projba. Dotting that difference with b gives zero directly. Together the two components form a right triangle, so their squared magnitudes add to |a|2. This gives an independent length check after the algebraic perpendicularity check.
Scaling b by a nonzero number does not change the vector projection: the line onto which we project is unchanged. Reversing b does reverse the signed scalar resolute, because its positive measuring direction has changed. This is a useful way to remember why the two kinds of resolute must not be confused.
Worked example 2: Find the nearest point on a line
A line passes through A = (1, 0, 2) in direction d = (2, 1, −2). Find the point H on the line nearest to P = (4, 4, 1), and the distance PH.
The displacement from A to P is w = (3, 4, −1). Its projection coefficient along d is (w · d)/|d|2 = (6 + 4 + 2)/9 = 4/3. Hence H = A + (4/3)d = (11/3, 4/3, −2/3).
The displacement from H to P is (1/3, 8/3, 5/3). It is perpendicular to d, since 2/3 + 8/3 − 10/3 = 0. Its length is √(90/9) = √10. Independently, |w|2 − |projdw|2 = 26 − 16 = 10. Any other point on the line adds a nonzero perpendicular-triangle leg, increasing the distance.
4. Turn a geometric claim into a dot-product statement
A vector proof begins with a diagram and clear definitions, but its justification comes from identities that work for every configuration satisfying the assumptions. Choose a convenient origin, label position vectors, and express the relevant sides or diagonals as differences. Then translate the desired geometric conclusion: perpendicularity becomes a zero dot product; equal length becomes equal squared magnitude; parallelism becomes a scalar multiple.
Do not use the desired conclusion as an assumption. For example, drawing a rhombus with perpendicular diagonals does not establish that all rhombi have that property. Nor is checking one numerical rhombus a general proof. Use arbitrary side vectors of equal magnitude and show that the condition forces the conclusion.
Squared magnitudes often keep proofs cleaner than square roots. Equal non-negative lengths have equal squares, and equal squares of non-negative lengths imply equal lengths. After completing the algebra, explicitly reconnect the result to the geometric claim. A line of symbols ending in zero is incomplete if the reader has to guess which objects were proved perpendicular.
Worked example 3: Prove the diagonals of a rhombus are perpendicular
Let adjacent side vectors of a non-degenerate parallelogram be u and v. Its diagonal directions can be taken as u + v and u − v. For a rhombus, the side lengths are equal, so |u| = |v|.
Expand their dot product: (u + v) · (u − v) = |u|2 − u · v + v · u − |v|2 = 0. The mixed terms cancel by commutativity, and the squared lengths cancel by the rhombus condition.
The non-degenerate assumption ensures the diagonals have genuine directions. Therefore the two diagonals are perpendicular. The argument also proves a converse within non-degenerate parallelograms: perpendicular diagonals force equal adjacent side lengths, hence a rhombus. The conclusion depends on the parallelogram structure, not merely on two arbitrary crossing segments.
5. Technology and non-routine checks
For a geometric proof, record exactly which assumptions justify each cancellation. Equal side lengths permit cancellation of squared magnitudes; perpendicular sides permit removal of a mixed dot product; independence permits comparison of vector coefficients. These are different hypotheses. Swapping one for another because two diagrams look similar can turn a valid special-case argument into an invalid general claim.
A symbolic tool can expand a dot product, solve a perpendicularity condition or simplify a projection. Keep the mathematical setup visible. Enter the projection as a scalar coefficient multiplying the complete direction vector, with brackets around each component. Then check both the parallel component and the perpendicular remainder rather than trusting one output.
For a point-to-line distance investigation, write H(t) = A + td and form |P − H(t)|2. This is a quadratic with positive leading coefficient |d|2. Completing the square gives its unique minimum at t = [(P − A) · d]/|d|2. This confirms the projection method without relying on the appearance of a diagram.
Exact arithmetic also protects geometric conclusions. If an exact perpendicularity calculation gives zero, do not replace it by a rounded decimal angle and then argue that the angle is only approximately right. Conversely, a numerical dot product close to zero is evidence of near-perpendicularity, not an exact proof, unless the original data or a justified tolerance supports that interpretation.
If the allowed set is a finite segment, this unrestricted minimum may lie beyond an endpoint. Then compare the permitted endpoint distances instead. A formula for a whole line does not automatically solve a constrained segment problem. Interpreting the domain is part of the mathematics.
Practise, then check
- Find the angle between (1, 1, 0) and (1, 0, 1).
- Determine k so that (2, k, −1) and (3, −2, 4) are perpendicular.
- Find the signed scalar and vector resolutes of (1, −3, 2) in the direction (0, 2, 0).
- Resolve (4, 1) into components parallel and perpendicular to (1, 2).
- Prove that the diagonals of a rectangle have equal length using perpendicular adjacent side vectors.
- If |a + b| = |a − b| for nonzero vectors, what follows about their angle?
- For real t, determine when (1, t) and (t, −2) are perpendicular and whether they can ever be parallel.
Fully reasoned answers
- The dot product is 1 and both lengths are √2. Therefore cos θ = 1/2, so θ = π/3 or 60°. Both vectors are nonzero, making the angle meaningful.
- The dot product is 6 − 2k − 4 = 2 − 2k. Setting it to zero gives k = 1. Neither vector becomes zero, so this is a valid perpendicularity condition.
- The dot product is −6 and the direction magnitude is 2. The scalar resolute is −3. The vector resolute is (−6/4)(0, 2, 0) = (0, −3, 0). Its magnitude is 3, while its direction is opposite the stated positive direction.
- The dot product is 6 and direction squared magnitude is 5. The parallel part is (6/5, 12/5); subtraction gives (14/5, −7/5). Their sum is (4, 1), and the remainder dotted with (1, 2) is zero.
- For adjacent vectors u · v = 0. Thus |u + v|2 = |u|2 + |v|2 = |u − v|2. Non-negative lengths therefore agree. These sum and difference vectors describe the diagonals.
- Squaring and expanding gives a difference of 4a · b = 0. Since both vectors are nonzero, cos θ = 0 and their angle is π/2. The conclusion would need qualification if a zero vector were allowed.
- The dot product is t − 2t = −t, so perpendicularity occurs at t = 0. Parallelism would require the two-dimensional component determinant −2 − t2 to be zero, impossible for real t. Equivalently, the scalar-multiple equations lead to t2 = −2.
Continue to lines and shortest distances, or revisit vector algebra.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4. Scope follows the VCAA Mathematics Study Design 2023–2027, Specialist Mathematics Area of Study 5: Space and measurement, with Area of Study 1 proof connections as applicable, and the current Specialist Mathematics FAQ. Unit 2 vector and conic foundations are retained where required. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 4C–4F, pp. 167–186. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
