Loav · VCE Specialist Mathematics · Units 3 & 4
S34-V04 · Vectors · Planes in space
1. The cross product supplies a perpendicular direction
The dot product of two vectors is a scalar. The cross product of two three-dimensional vectors is a vector perpendicular to both. For nonparallel vectors u and v, its direction is selected by the right-hand rule: curl from u towards v through the smaller angle, and the thumb indicates u × v. Its magnitude is |u||v| sin θ.
This magnitude is the area of the parallelogram spanned by the vectors. Half the magnitude gives the area of the corresponding triangle. Parallel vectors, or a pair containing the zero vector, have cross product zero. In those cases the result has no normal direction, so it cannot define a plane normal.
The determinant mnemonic arranges the coordinate unit vectors over the two component rows. Expand along the first row with signs plus, minus, plus. The middle minus sign is easy to lose; the component formula above already includes it.
Reversing the order changes the sign: v × u = −(u × v). Both directions are perpendicular to the same plane, so either can serve as a normal when the plane equation is adjusted consistently. Always verify a computed cross product by dotting it with both original vectors. This catches arithmetic errors independently of the determinant expansion.
2. Three connected ways to describe a plane
A plane can be specified by a point A and two nonparallel directions u and v lying in it. Its vector-parametric form is r = a + su + tv, where s and t independently range over all real numbers. Two independent directions are necessary; parallel directions would trace only a line.
Expanding the coordinates gives the parametric equations x = a₁ + su₁ + tv₁, y = a₂ + su₂ + tv₂ and z = a₃ + su₃ + tv₃. A parameter pair selects a point. Unlike a line, a plane needs two free parameters. Restricting these parameters may describe only a bounded patch, so distinguish the whole plane from a finite physical surface.
A second vector description uses a nonzero normal n. Every displacement from A to another plane point is perpendicular to n, giving (r − a) · n = 0, or r · n = a · n. Expanding this gives the Cartesian form n₁x + n₂y + n₃z = d, where d = a · n.
Thus the Cartesian coefficients give a normal, not a direction within the plane. Multiplying the entire Cartesian equation by any nonzero constant preserves the plane. Multiplying only the left side changes its location unless the right side happens to be zero. A missing variable means the plane is parallel to that coordinate direction, not that the variable equals zero.
Worked example 1: Construct every plane form
A plane contains A = (1, 0, −1) and has directions u = (1, 1, 0) and v = (0, 1, 2). Find its equations and the area of the parallelogram defined by those directions.
The cross product is n = (2, −2, 1), which is nonzero. Its dot products with u and v are 2 − 2 = 0 and −2 + 2 = 0. The vector-parametric form is r = (1, 0, −1) + s(1, 1, 0) + t(0, 1, 2).
The parametric equations are x = 1 + s, y = s + t and z = −1 + 2t. The normal-vector form is [r − (1, 0, −1)] · (2, −2, 1) = 0. Since a · n = 1, the Cartesian equation is 2x − 2y + z = 1. Substitution of the parametric coordinates makes the left side identically one.
The parallelogram area is |n| = √(4 + 4 + 1) = 3 square units. The associated triangle area is 3/2. The cross product therefore encodes both a perpendicular direction and a geometric size.
3. Line–plane intersections have three possibilities
Substitute a line r = a + td into a plane r · n = c. The resulting equation is a · n + t(d · n) = c. If d · n is nonzero, there is exactly one parameter and therefore one intersection point.
If d · n = 0, the line direction is parallel to the plane. Now inspect its starting point. When a · n = c, every point on the line is in the plane. Otherwise no point is in the plane. Do not divide by the zero coefficient and treat a resulting error as a geometric conclusion.
The acute angle α between a line and a plane satisfies sin α = |d · n|/(|d||n|). The sine appears because the line–plane angle complements the acute angle between the line direction and the normal. A line parallel to the plane gives α = 0; a normal line gives α = π/2.
4. Two planes: intersection, coincidence or separation
Compare the nonzero normals of two planes. If they are not parallel, the planes intersect in one straight line. A direction for that line is their cross product, because it is perpendicular to both normals and therefore lies along both planes. Find one point satisfying both Cartesian equations, then combine that point with the direction.
If the normals are parallel, first scale the equations to use the same normal. Equal right sides then mean coincident planes; unequal right sides mean distinct parallel planes. The smaller angle between planes is the smaller angle between their normals: cos φ = |n₁ · n₂|/(|n₁||n₂|). The absolute value removes the arbitrary choice of normal orientation.
Worked example 2: Find the line and angle between two planes
Let Π₁ be x + y + z = 3 and Π₂ be x − y + 2z = 2. Their normals are (1, 1, 1) and (1, −1, 2), whose cross product is (3, −1, −2).
Setting z = 1 gives x + y = 2 and x − y = 0, so (1, 1, 1) belongs to both planes. Using the negative of the computed direction is equally valid: r = (1, 1, 1) + t(−3, 1, 2). Substitution gives 3 and 2 in the two plane equations for every real t.
The dot product of the normals is 2; their lengths are √3 and √6. Thus φ = cos−1(2/√18), approximately 61.9°. This is the smaller plane angle. A perspective sketch can help display the intersection, but the simultaneous equations and normal calculations establish it exactly.
5. Distance is measured along a unit normal
For a point P with position vector p and plane r · n = c, the signed normal displacement is (p · n − c)/|n|. Its absolute value is the point-to-plane distance. The numerator alone is not a distance unless the normal is a unit vector. Its sign depends on normal orientation.
Perpendicular foot: h = p − [(p · n − c)/|n|2]n
For parallel planes written with the same normal, r · n = c₁ and r · n = c₂, their distance is |c₂ − c₁|/|n|. Normalise the coefficient scale before subtracting constants. Nonparallel infinite planes intersect, so their minimum separation is zero.
Worked example 3: A perpendicular foot and parallel-plane distance
Find the distance and perpendicular foot from P = (3, −1, 4) to 2x − y + 2z = 5. Then find this plane's distance from 2x − y + 2z = −4.
Here n = (2, −1, 2), |n| = 3 and p · n − 5 = 6 + 1 + 8 − 5 = 10. The point distance is 10/3. The foot is P − (10/9)(2, −1, 2) = (7/9, 1/9, 16/9).
Substituting the foot gives 14/9 − 1/9 + 32/9 = 5. Its displacement from P is parallel to the normal, confirming perpendicularity. The parallel planes already use the same coefficients, so their separation is |5 − (−4)|/3 = 3. The answer is a length, not the unnormalised constant difference nine.
6. Verification and geometric interpretation
For a symbolic or numerical check, substitute a proposed intersection line into both plane equations and simplify. A correct answer makes both expressions constant for every parameter, not merely at one sampled point. Check a plane normal against both spanning directions and check any proposed perpendicular foot against the original plane.
When three non-collinear points specify a single plane, subtract one chosen point from the other two to form spanning vectors. Their cross product gives a normal, and the chosen point fixes the constant. If the cross product vanishes, the points are collinear or repeated and do not determine a unique plane. This is a degeneracy in the supplied data, rather than a reason to divide by a zero normal magnitude.
To recover a parametrisation from a Cartesian equation, choose two coordinates freely and solve for a coordinate with a nonzero coefficient. For example, if the z-coefficient is nonzero, set x = s and y = t, then solve for z. If that coefficient is zero, choose a different coordinate to solve for. This makes the two independent degrees of freedom visible and avoids treating a fixed-coordinate plane as exceptional.
In an examination response, match the final representation to the command. A Cartesian equation is not the requested parametric answer, even if it represents the correct set. Conversely, a list of two direction vectors does not identify a plane until a point is supplied. Showing the conversion between representations makes both the location and the orientation explicit.
A useful investigation varies one plane's constant while leaving its normal fixed. The plane translates parallel to itself; its orientation does not change. The distance moved equals the absolute constant change divided by the normal magnitude. This explains why different coefficient scales must be reconciled before comparing constants. The geometry here concerns one plane or a pair of planes.
Practise, then check
- Find (1, 2, 0) × (0, 1, 3) and the spanned parallelogram area.
- Give Cartesian and vector-parametric equations for the plane through (2, −1, 0) with normal (1, 2, −1).
- Intersect r = (1, 0, 0) + t(1, 2, 1) with x + y + z = 5.
- Classify the relationship of r = (1, 1, 0) + t(1, −1, 0) to each of x + y = 2 and x + y = 3.
- Find the distance between 2x − y + 2z = 1 and 4x − 2y + 4z = 10.
- Find the intersection line and angle of planes x = 1 and y = 2.
- Find the acute angle between a line of direction (1, 1, 1) and the plane z = 0.
Fully reasoned answers
- The cross product is (6, −3, 1). Dotting it with each original vector gives zero. Its magnitude is √(36 + 9 + 1) = √46, the parallelogram area; reversing the product changes direction but not area.
- The normal form gives (x − 2) + 2(y + 1) − z = 0, so x + 2y − z = 0. One parametrisation is r = (2, −1, 0) + s(−2, 1, 0) + t(1, 0, 1). Both independent directions have zero dot product with the normal.
- Substitution gives 1 + t + 2t + t = 5, hence t = 1. The intersection point is (2, 2, 1), and its coordinate sum is five.
- Every point on the line has x + y = (1 + t) + (1 − t) = 2. Therefore the line lies in the first plane and is parallel to, but disjoint from, the second.
- Divide the second equation by two to obtain 2x − y + 2z = 5. The common normal has magnitude three, so the distance is |5 − 1|/3 = 4/3. Subtracting ten and one before rescaling would be wrong.
- Both fixed coordinates hold on r = (1, 2, 0) + t(0, 0, 1). The normals (1, 0, 0) and (0, 1, 0) are perpendicular, so the plane angle is π/2.
- A normal is (0, 0, 1). Therefore sin α = 1/√3 and α = sin−1(1/√3), approximately 35.3°. Using inverse cosine would give the complementary angle to the normal.
Continue to parametric curves, or review line equations.
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 5C–5E, pp. 212–229. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
