Loav · VCE Specialist Mathematics · Units 3 & 4
S34-V01 · Vectors · Foundations
1. A vector records a directed change
A scalar describes an amount. A vector describes an amount together with a direction. The displacement from a starting point to a finishing point is a vector: travelling six units east and three units south cannot be described adequately by giving only the distance between the points. In a diagram, an arrow represents the vector, with its length representing magnitude and its arrowhead indicating direction.
Two arrows represent the same free vector when they have equal length and the same direction, even if they start at different places. A position vector is more specific: it starts at a chosen origin and ends at a point. Write a for the position vector of A. If B has position vector b, the displacement from A to B is b − a. The order matters. Reversing the journey gives a − b, the negative of the original vector.
Use bold letters for vectors and ordinary letters for scalars. Here a coordinate tuple such as (2, −1, 3) represents vector components when it appears in a vector equation, and a point when explicitly attached to a point label. A zero vector has every component zero. It has magnitude zero and no defined direction, so it cannot be converted into a unit vector.
2. Components turn geometry into algebra
The unit coordinate vectors i, j and k point along the positive x-, y- and z-axes. They each have length one and are mutually perpendicular. Thus (x, y, z) means xi + yj + zk. A negative component means movement opposite the corresponding positive axis. Components are signed quantities; a magnitude is non-negative.
Add vectors component by component. Geometrically, place the second arrow's tail at the first arrow's head. The arrow from the original start to the final finish represents their sum. The same sum is the diagonal of the parallelogram constructed from the two vectors at a common origin. Subtraction is addition of the opposite vector. Multiplying by a positive scalar changes length but preserves direction; multiplying by a negative scalar also reverses direction.
For a ≠ 0, the unit vector along a is a/|a|.
The magnitude formula follows by applying Pythagoras to perpendicular components. It does not say that magnitudes add when vectors add. Two opposite displacements can have a zero resultant despite each having positive magnitude. Also, |λa| = |λ||a|: the absolute value on the scalar is essential because reversing an arrow cannot make its length negative.
In a plane, a vector of magnitude R at angle θ anticlockwise from the positive x-axis has components (R cos θ, R sin θ). This assumes the angle is measured from that axis. If a question gives a bearing measured clockwise from north, first translate the direction into the coordinate convention or resolve directly using a sketch. Check signs against the quadrant before calculating decimals.
Worked example 1: Separate position, displacement and direction
A has coordinates (−2, 1, 3) and B has coordinates (4, −2, 5). Find the displacement from A to B and the point P five units from A in the direction of B.
The displacement is d = (4, −2, 5) − (−2, 1, 3) = (6, −3, 2). Its length is √(36 + 9 + 4) = 7, so its unit direction is (6, −3, 2)/7. A displacement of length five in that direction is 5d/7, not 5d.
Add this displacement to the starting position: p = (−2, 1, 3) + (30/7, −15/7, 10/7) = (16/7, −8/7, 31/7). Since 0 < 5/7 < 1, P lies inside segment AB. Checking |p − a| = 5 confirms the requested distance; checking that p − a is a positive multiple of d confirms the direction.
3. Linear combinations and dependence
A linear combination of vectors is a sum of scalar multiples, such as 3u − 2v. The coefficients can be any real numbers unless the problem imposes restrictions. To determine coefficients, equate corresponding components and solve the resulting equations. Every component must work. Satisfying the first two equations of a three-dimensional vector equation is only a candidate solution.
Vectors are linearly dependent when a combination of them equals the zero vector with at least one coefficient nonzero. They are linearly independent when the only combination giving zero has all coefficients zero. The nonzero-coefficient requirement matters: zero times each vector always produces zero and therefore cannot distinguish dependence from independence.
For two nonzero vectors, dependence means one is a scalar multiple of the other. They point along parallel directions, possibly opposite directions. Two nonparallel vectors are independent and generate a plane of possible linear combinations. Three vectors can be dependent without being pairwise parallel: one may lie in the plane generated by the other two. These are geometric tests in ordinary two- and three-dimensional space, rather than a separate abstract theory.
A collection containing the zero vector is dependent: give that vector coefficient one and the others coefficient zero. When checking multiples, avoid dividing by a component that might be zero. Solve u = λv componentwise instead. A zero component in one vector may immediately rule out a scalar multiple, or may impose no new condition if both corresponding components are zero.
Worked example 2: Dependent does not mean every pair is parallel
Let u = (1, 2, −1), v = (2, −1, 3) and w = (7, 4, 3). Express w using u and v, then interpret the result.
Set αu + βv = w. The first two components give α + 2β = 7 and 2α − β = 4. Solving gives α = 3 and β = 2. The third component is −3 + 6 = 3, so the candidate satisfies the entire vector equation.
Thus w − 3u − 2v = 0, a nontrivial dependence relation. Yet u and v are not multiples: their first components would require the multiplier two, whereas their second components would then disagree. The three vectors lie in one plane through the origin. Their dependence does not imply that the corresponding endpoints lie on one straight line.
4. Collinearity concerns differences between points
Points A, B and C are collinear when their locations lie on one straight line. For distinct A and B, test whether c − a = λ(b − a) for some real λ. This compares two displacements sharing the same starting point. Testing whether a, b and c are scalar multiples would incorrectly require their line to pass through the origin.
The expression p = a + t(b − a) also locates a point along the line. When t = 0 it gives A; when t = 1 it gives B. Values between zero and one describe the segment. Values above one pass beyond B, while negative values go beyond A in the opposite direction. This parameter interpretation is often safer than memorising an internal-division formula.
If P divides AB internally in ratio AP:PB = m:n, where m and n are positive, then t = m/(m + n). Therefore p = [na + mb]/(m + n). The weights appear opposite the corresponding endpoint labels because moving a larger fraction of the way towards B makes the coefficient of b larger. A midpoint has equal weights.
Worked example 3: Internal and external division
Let A = (1, −2, 2) and B = (6, 3, −3). Find Q with AQ:QB = 2:3 internally, and R defined by displacement AR = (3/2)AB.
Here AB = (5, 5, −5). For Q, the fraction of the whole segment is 2/5, giving Q = (1, −2, 2) + (2, 2, −2) = (3, 0, 0). The lengths AQ and QB are respectively (2/5)|AB| and (3/5)|AB|, so their ratio really is 2:3.
For R, use the stated directed relationship: R = (1, −2, 2) + (15/2, 15/2, −15/2) = (17/2, 11/2, −11/2). The parameter 3/2 exceeds one, so R lies beyond B. In ordinary positive lengths AR:RB = 3:1, because RB is half of AB. An external point should not be forced into an internal weighted average.
5. Check a result algebraically and geometrically
Before using technology, predict what a reasonable answer must do. A unit vector must have squared magnitude one. A point inside a segment must be expressible with a parameter between zero and one. A claimed dependence relation must produce zero in every component. These checks diagnose meaning, not merely arithmetic.
With a computer algebra system, solve the coefficient equations simultaneously, then substitute the result into the original vector equation. For a parameter-dependent problem, retain special parameter values rather than cancelling an expression that could be zero. A numerical graph can suggest dependence, but apparent overlap in a perspective drawing is not proof: a three-dimensional diagram flattens depth onto a page.
A useful extension asks when u + 2v and 3u + kv are parallel, assuming u and v are independent. Independence allows their coefficients to be compared. The first coefficient forces a multiplier three, so the second forces k = 6. Without the independence assumption, this coefficient argument would not be justified.
Practise, then check
- For a = (2, −1, 3) and b = (−1, 4, 0), find 2a − b and |a|.
- If (p + 1, 2q, −3) = (4, 6, r), determine p, q and r.
- Find the unit vector in the direction (−2, 3, 6), and the vector of magnitude 21 in that direction.
- Are A = (−1, 2, 0), B = (2, 4, −2) and C = (5, 6, −4) collinear? Locate B relative to A and C.
- Show that (1, 0, 2), (0, 1, −1) and (2, 3, 1) are dependent, but the first two are independent.
- Resolve a vector of magnitude 12 at 150° anticlockwise from the positive x-axis.
- For independent u and v, determine when u + 2v and 3u + kv are parallel. Explain the role of independence.
Fully reasoned answers
- Doubling first gives (4, −2, 6), so subtraction gives (5, −6, 6). The magnitude is √(4 + 1 + 9) = √14. The negative y-component contributes a positive square.
- Vector equality means component equality: p + 1 = 4, 2q = 6 and r = −3. Hence p = 3, q = 3 and r = −3. A single scalar equality would not contain all the information.
- The original magnitude is √49 = 7, giving (−2/7, 3/7, 6/7). Multiplying this unit vector by 21 gives (−6, 9, 18). Its squared magnitude is 441 and its multiplier from the original is positive.
- AB = (3, 2, −2) and AC = (6, 4, −4) = 2AB. Hence the points are collinear and B is the midpoint of AC. This proves the location without relying on a perspective sketch.
- Twice the first plus three times the second equals (2, 3, 1). Moving this third vector to the other side gives a nontrivial zero combination. For the first two alone, their x- and y-components force both coefficients to zero, proving independence.
- The components are (12 cos 150°, 12 sin 150°) = (−6√3, 6). The signs match quadrant two, and the squared magnitude is 108 + 36 = 144.
- Write 3u + kv = λ(u + 2v). Independence forces 3 = λ and k = 2λ, so k = 6. The vectors then differ by a factor three. Without independence, unequal coefficient pairs could represent the same vector.
Next: dot products and projections add angle and perpendicularity to this 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 4A–4B, pp. 144–166; 4E, pp. 176–178. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
