Loav · VCE Specialist Mathematics · Units 3 & 4
S34-V03 · Vectors · Lines in space
1. A line is a point plus a direction
A straight line is determined by one point A on it and one nonzero direction vector d. Starting at A, all positive and negative scalar multiples of d trace the complete line. If a is the position vector of A, write r = a + td, where t ranges over the real numbers. The parameter is a coordinate along the line, not automatically time or distance.
The direction vector must be nonzero. Otherwise every parameter gives the same point. Multiplying d by a nonzero scalar does not change the geometric line, provided the parameter still ranges over every real number. It changes the parameter scale and may reverse increasing-parameter orientation. Similarly, choosing another starting point on the line gives a different-looking equation for the same set of points.
If two distinct points A and B are given, use d = b − a. Then t = 0 locates A and t = 1 locates B. Restricting 0 ≤ t ≤ 1 describes the segment AB. Restricting t ≥ 0 describes a ray from A towards B. Always preserve a stated parameter restriction when deciding whether a point belongs to the required object.
2. Translate forms without dividing by zero
Writing the vector equation componentwise gives parametric equations x = a₁ + td₁, y = a₂ + td₂ and z = a₃ + td₃. Each component uses the same parameter because the coordinates belong to one point. Where all direction components are nonzero, eliminate t to obtain the symmetric Cartesian form (x − a₁)/d₁ = (y − a₂)/d₂ = (z − a₃)/d₃.
A zero direction component says that the corresponding coordinate is constant. It does not justify a fraction with denominator zero. For example, d₃ = 0 gives z = a₃, alongside the relation obtained from the remaining components. In a plane, a horizontal line has constant y and a vertical line has constant x. Neither requires an artificial infinite slope.
A Cartesian description of a three-dimensional line usually needs two independent conditions: one equation alone generally describes a plane. When converting back to vector form, identify one point satisfying both conditions and a direction preserving both. An equation is complete only when it includes enough information to describe the intended set.
Worked example 1: A line with a fixed coordinate
Find vector, parametric and Cartesian descriptions of the line through A = (2, −1, 4) and B = (5, 1, 4).
The direction B − A is (3, 2, 0). A vector equation is r = (2, −1, 4) + t(3, 2, 0), t ∈ ℝ. Its parametric equations are x = 2 + 3t, y = −1 + 2t and z = 4.
Eliminating t from the first two coordinates gives (x − 2)/3 = (y + 1)/2, together with z = 4. The fixed z-coordinate is essential. Equivalently, use 2x − 3y = 7 and z = 4. Substituting both A and B checks the Cartesian conditions. The direction lies parallel to the xy-plane, although the line itself lies at height four.
3. Classify two lines systematically
First compare the direction vectors. If they are parallel, the lines either coincide or are distinct parallel lines. To distinguish these possibilities, test whether a point on one line lies on the other. Matching directions alone does not establish coincidence because parallel lines can be translated apart.
If the directions are not parallel, solve for an intersection. Use separate parameters, such as s for the first line and t for the second. Equality of positions gives one equation per coordinate. In two dimensions, nonparallel complete lines intersect. In three dimensions, they may instead be skew: nonparallel and nonintersecting. Skew lines do not lie in a common plane.
With three component equations and two parameters, solve any convenient pair, then substitute into the remaining equation. A contradiction means no intersection. Do not discard that final equation because the first two gave neat values. Conversely, a common point requires every coordinate to agree for the same parameter pair.
Using the same parameter on both lines would impose an extra condition that the chosen parameter values agree. That is relevant when both parameters genuinely represent the same time and you are testing a collision. It is generally wrong for a geometric line intersection, where the two parameter scales have no reason to match.
Worked example 2: An intersection and a nearby skew case
Consider L: r = (1, 2, −1) + s(2, −1, 3) and M: r = (5, −1, 4) + t(0, 1, 1). Classify the lines and find their intersection if one exists.
The directions are not multiples. Equating x-coordinates gives 1 + 2s = 5, so s = 2. The y-equation becomes 2 − 2 = −1 + t, so t = 1. The z-equation is −1 + 3(2) = 4 + 1, which is true. Therefore the lines intersect at (5, 0, 5).
Now replace M's starting point by (5, −1, 5), keeping its direction. The x- and y-equations still give s = 2 and t = 1, but the z-values become five and six. There is no intersection. Because the directions remain nonparallel, the modified lines are skew. A small change in one coordinate has changed the geometry, even though their projections onto the xy-plane still cross.
4. Angles belong to direction vectors
The smaller angle φ between two lines is determined by their nonzero direction vectors d and e: cos φ = |d · e|/(|d||e|), with 0 ≤ φ ≤ π/2. The absolute value makes the result independent of which orientation was chosen for either line. Two skew lines still have an angle between their directions, even though there is no crossing point at which to draw it.
A point-to-line distance is the length of a perpendicular displacement. From a line point A to an external point P, form w = P − A. Remove its projection along the line: w − [(w · d)/|d|2]d. The magnitude of the remainder is the required distance. The projection coefficient also locates the nearest line point.
For two distinct parallel lines, use any point on one line and find its distance to the other. For intersecting or coincident lines, the shortest distance is zero. For skew lines, both endpoints of the shortest joining segment must be determined. A distance between two arbitrary starting points is generally too large.
Worked example 3: Find the shortest segment between skew lines
Let L have points P(s) = (s, s, 0), and let M have points Q(t) = (1 + t, 0, 2 + t). The direction vectors are d = (1, 1, 0) and e = (1, 0, 1). Find the shortest distance.
The joining vector is w = Q − P = (1 + t − s, −s, 2 + t). For a common perpendicular, require w · d = 1 + t − 2s = 0 and w · e = 3 + 2t − s = 0. Solving gives s = −1/3 and t = −5/3.
The endpoints are P = (−1/3, −1/3, 0) and Q = (−2/3, 0, 1/3). Their difference is (−1/3, 1/3, 1/3), whose length is √3/3. Its dot product with each direction is zero, confirming both conditions.
To justify minimality, any other joining vector differs by βe − αd. That difference is perpendicular to the common-perpendicular vector. Pythagoras therefore adds a non-negative squared length to the value already found. Because the directions are nonparallel, the added vector vanishes only when α = β = 0. The calculated pair uniquely minimises distance.
5. A reliable technology workflow
Parameter values are not generally distances along a line. A change of parameter Δt produces displacement Δtd, whose length is |Δt||d|. Only a unit direction vector makes the absolute parameter change equal to distance. This distinction matters when a question compares segment lengths or interprets a parameter as elapsed time. If the direction vector represents velocity, its magnitude supplies the conversion from time to distance for that straight motion.
A useful coincidence check also compares whole sets rather than identical formulas. Show that one starting point lies on the other line and that their nonzero directions are parallel. These two facts establish every point in common, even when the parameter names and scales differ.
For an intersection, ask a symbolic solver to solve every component equation simultaneously, with distinct parameter names. If it returns no solution, separately check whether the directions are parallel before naming the relationship. A failed intersection calculation alone cannot distinguish skew from parallel lines.
For shortest distance, solve the two perpendicularity equations or minimise the squared distance. Squaring removes the square root without changing the location of the minimum, since distance is non-negative. Verify the final connecting vector against both directions. A three-dimensional graph is useful for orientation, but it is not a replacement for the exact algebra.
A non-routine variation replaces complete lines by two finite tracks. The common-perpendicular solution is valid only if both parameters lie inside their allowed intervals. Otherwise the minimum occurs with at least one point at a boundary, so test endpoint-to-segment distances. Similarly, a geometric crossing of complete lines may lie outside the two actual segments. Always state which object your final answer concerns.
Practise, then check
- Write vector and Cartesian equations for the line through (−1, 2, 3) with direction (0, 2, −1).
- Classify L: (0, 0, 0) + s(1, 2, 0) and M: (0, 1, 0) + t(2, 4, 0).
- Does (3, 5, 7) + t(2, 2, 2) describe the same line as (1, 3, 5) + s(1, 1, 1)?
- Classify L: (s, 0, 0) and M: (0, t, 1).
- Find the smaller angle between directions (1, 1, 0) and (1, 0, 1).
- Find the distance from (2, 3, 0) to the x-axis and locate its nearest point.
- The segments P(s) = (s, 0, 0) and Q(t) = (2, t, 0) have 0 ≤ s ≤ 1 and −1 ≤ t ≤ 1. Do they intersect? Find their shortest distance.
Fully reasoned answers
- The vector equation is r = (−1, 2, 3) + t(0, 2, −1). Cartesian conditions are x = −1 and (y − 2)/2 = (z − 3)/(−1), equivalently, x = −1 together with y + 2z = 8. The zero x-direction requires a fixed coordinate, not division by zero.
- The directions are multiples. The point (0, 1, 0) is not on L: x = 0 requires s = 0, but then y = 0. Hence the lines are distinct and parallel.
- Yes. The second starting point plus 2(1, 1, 1) is (3, 5, 7), and the first direction is twice the second. The parameter relationship is s = 2 + 2t, which covers every real s.
- They are nonparallel, but one line has z = 0 and the other z = 1. No intersection is possible, so they are skew. Their shortest distance is one, joining the points (0, 0, 0) and (0, 0, 1).
- The dot product is 1 and the magnitude product is 2. Thus cos φ = 1/2 and φ = 60°. The directions are not parallel, and the result is the smaller line angle.
- The nearest point is (2, 0, 0), obtained by retaining the x-coordinate. The remaining displacement (0, 3, 0) is perpendicular to the axis and has magnitude three.
- The complete lines cross at (2, 0, 0), requiring s = 2, outside the first segment. Thus the segments do not intersect. Their nearest endpoints/points are P(1) = (1, 0, 0) and Q(0) = (2, 0, 0), giving distance one; any other choice increases (2 − s)2 + t2.
Next: cross products and planes. Review projections if shortest-distance reasoning feels unfamiliar.
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 5A–5B, pp. 198–211; 5E, pp. 223–229. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
