MTH1020 · Analysis of Change
Vectors: Cross Product, Lines & Planes
Week 4 of Monash MTH1020 Analysis of Change extends vectors into three-dimensional geometry. It defines the cross (vector) product a × b — a vector perpendicular to both inputs, with magnitude |a||b|sin θ and a right-hand-rule direction — and uses it for areas and normals. It then describes lines by r(t) = a + td and planes by (r − a) · n = 0, and solves the intersection, angle and distance problems (point–line, line–line, point–plane). These multi-step constructions are staple mid-semester-test and final-exam questions, and correct components plus a clear method earn the marks.
What this chapter covers
- 01Cross product a × b: perpendicular to both a and b, magnitude |a||b|sin θ, right-handed set
- 02Component formula a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁); anticommutative a × b = −(b × a)
- 03Area applications: |a × b| = parallelogram area; triangle area = ½|AB × AC|
- 04Lines: vector equation r(t) = a + td; through two points r(t) = (1−t)a + tb; Cartesian form in ℝ³
- 05Intersections of lines: parallel/coincident/single-point/skew; test d₁ = k d₂ then solve r₁(t) = r₂(u)
- 06Planes: normal vector n; vector equation (r − a) · n = 0 ⇔ r · n = a · n; Cartesian n₁x + n₂y + n₃z = k
- 07Normal to a plane through three points: n = AB × AC; angle between planes via their normals
- 08Distances: point–plane d = |PQ · n̂|; between skew lines d = |(a₂ − a₁) · n̂| with n̂ = (d₁ × d₂)/|d₁ × d₂|
Cross product: triangle area and the plane through three points
- +1Form the edge vectors from A. AB = B − A = (2−1, 1−0, 3−1) = (1, 1, 2) and AC = C − A = (0−1, 2−0, 2−1) = (−1, 2, 1).
- +1Cross product (component formula). AB × AC = (1·1 − 2·2, 2·(−1) − 1·1, 1·2 − 1·(−1)) = (1 − 4, −2 − 1, 2 + 1) = (−3, −3, 3).
- +1Area of the triangle. |AB × AC| = √((−3)² + (−3)² + 3²) = √(9 + 9 + 9) = √27 = 3√3, so Area = ½|AB × AC| = (3√3)/2 ≈ 2.60.
- +1Plane through the three points. A normal is n = AB × AC = (−3, −3, 3), or the simpler parallel (1, 1, −1). Using r · n = a · n with A(1, 0, 1): x + y − z = 1 + 0 − 1 = 0, so the plane is x + y − z = 0. (Check B: 2 + 1 − 3 = 0 ✓; C: 0 + 2 − 2 = 0 ✓.)
Key terms
- Cross (vector) product
- For 3-D vectors, a × b is perpendicular to both a and b, has magnitude |a||b|sin θ, and forms a right-handed set. Components: (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁).
- Anticommutativity
- The cross product satisfies a × b = −(b × a), so swapping the order reverses the result. If a ∥ b then a × b = 0.
- Vector equation of a line
- r(t) = a + td, where a is a point on the line and d its direction. Through two points, d = b − a gives r(t) = (1−t)a + tb.
- Normal vector (n)
- A vector perpendicular to a plane. A plane through A with normal n has (r − a) · n = 0; for three points, n = AB × AC.
- Cartesian equation of a plane
- n₁x + n₂y + n₃z = k, where (n₁, n₂, n₃) is the normal and k = a · n. The coefficients of x, y, z are the normal's components.
- Skew lines
- Two lines in ℝ³ that are neither parallel nor intersecting. Their shortest distance is d = |(a₂ − a₁) · n̂| with n̂ = (d₁ × d₂)/|d₁ × d₂|.
Vectors: Cross Product, Lines & Planes FAQ
When do I use the dot product versus the cross product?
Use the dot product when you want a number or an angle — a · b = |a||b|cos θ gives angles, perpendicularity (a · b = 0) and projections. Use the cross product when you want a vector perpendicular to two others, or an area — a × b has magnitude |a||b|sin θ, points normal to both, and gives parallelogram and triangle areas. A useful memory hook: dot involves cos and stays in any dimension; cross involves sin, needs 3-D, and produces a new vector.
How do I find the equation of a plane through three points?
Form two edge vectors from one point, say AB and AC, then take their cross product to get a normal n = AB × AC. Write the plane as r · n = a · n, which in coordinates is n₁x + n₂y + n₃z = k with k found by substituting the known point. Simplify the normal to small integers first, and always check the remaining two points satisfy your equation.
What does it mean for two lines to be skew?
Skew lines are lines in three dimensions that never meet and are not parallel — impossible in the plane, but common in ℝ³. You detect them by checking the direction vectors are not scalar multiples (not parallel) and then finding that r₁(t) = r₂(u) has no solution (they don't intersect). Their shortest distance uses the common perpendicular direction n̂ = (d₁ × d₂)/|d₁ × d₂|.
Can Sia help me set up 3-D geometry problems?
Yes. Sia can walk through a cross-product calculation term by term, show how the normal becomes the Cartesian coefficients of a plane, and lay out the method for point–plane and skew-line distances step by step. It explains the reasoning and checks your working on your own practice questions; it does not complete graded assessment for you, and Monash academic-integrity rules apply.
Exam move
This is the most multi-step vectors material, so drill each construction as a fixed recipe. For the cross product, practise the component formula until the sign pattern is automatic, and always sanity-check by confirming your result is perpendicular to both inputs (two dot products of zero). For planes, rehearse 'edge vectors → cross product → normal → r · n = a · n → simplify', and verify with the spare point. For distances, memorise the two formulas (point–plane d = |PQ · n̂|; skew lines with n̂ from d₁ × d₂) and draw a quick diagram each time. Because MST2 covers Weeks 1–4, this chapter is examined together with the earlier ones, and it recurs in the final — so over-learn the recipes and ask Sia to generate fresh point sets to practise on.
Working through Vectors: Cross Product, Lines & Planes in MTH1020? Sia is AskSia’s AI Mathematics tutor — ask any MTH1020 Vectors: Cross Product, Lines & Planes question and get a clear, step-by-step explanation grounded in how MTH1020 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
