Lines and planes is the single most examined strand of the vectors topic, which accounts for roughly two questions in every H2 Mathematics paper. Every later skill (feet of perpendiculars, angles, intersections, reflections) starts with writing down, and translating between, the equations of lines and planes.
The notes start with the vector and cartesian equations of a line and the three forms of a plane. They then cover the foot-of-perpendicular method, which gives distances and reflections from one calculation. They finish with the sin-versus-cos rule for angles and a systematic procedure for deciding how two lines, a line and a plane, or two planes are positioned.
What these notes cover
- •Vector equation of a line
- •Converting between vector and cartesian forms of a line
- •Three forms of a plane
- •Converting between parametric and scalar-product forms
- •Plane through three points
- •Foot of perpendicular from a point to a line
- •Foot of perpendicular from a point to a plane
- •Reflection of a point
- •Angles between lines and planes
- •Two lines: parallel, intersecting or skew
- •Line and plane; two planes
- •Common mistakes and exam technique
Key definitions
- Direction vector
- Any nonzero vector parallel to the line. It may be replaced by any nonzero scalar multiple of itself, so the equation of a line is not unique.
- Normal vector
- A vector perpendicular to the plane. Together with one point on the plane, it fixes the plane completely.
- Skew lines
- Lines that are not parallel and do not intersect. They exist only in three dimensions.
- Foot of perpendicular
- The point F on a line or plane where the perpendicular from a given point meets it. F is also the midpoint of the point and its mirror image.
- Angle between a line and a plane
- The angle θ between the line and its projection onto the plane, which is the ‘shadow’ the line casts on the plane.
Key formulas
Vector equation of a line
a is the position vector of any point on the line and d is any nonzero vector parallel to it. Always state λ ∈ ℝ, and use a different parameter for each line.
Cartesian equation of a line
This form is valid only when d₁, d₂ and d₃ are all nonzero. If a component is zero, that coordinate is constant and is written as a separate equation, e.g. y = a₂. Never divide by 0.
Plane — scalar-product form
b and c are two non-parallel direction vectors in the plane and a is any point on it. Expanding with r = (x, y, z) gives the cartesian form ax + by + cz = d, where (a, b, c) is the normal.
Perpendicular distance from a point to a plane
The plane is π: r · n = D and p is the position vector of P. Keep the modulus: the dot product can be negative, but a distance cannot.
Reflection of a point
p′ = 2f − p
f is the position vector of the foot of perpendicular from P, which is the midpoint of P and its image P′. The same formula works for reflection in a line and reflection in a plane.
Angle between a line and a plane
d is the line's direction and n is the plane's normal. Two lines (d₁, d₂) and two planes (n₁, n₂) use cos θ instead. The modulus guarantees the acute angle.
Worked example
The plane π has equation r · (1, 2, −2) = 10 and the point P has coordinates (3, 1, 2). Find (i) the foot of perpendicular F from P to π, (ii) the perpendicular distance from P to π, and (iii) the point P′, the reflection of P in π.
- 1The line through P perpendicular to π is parallel to the normal n = (1, 2, −2), so ℓ: r = (3, 1, 2) + t(1, 2, −2), t ∈ ℝ.
- 2Substitute the general point (3 + t, 1 + 2t, 2 − 2t) into the plane equation: (3 + t) + 2(1 + 2t) − 2(2 − 2t) = 10.
- 3Expand: 3 + t + 2 + 4t − 4 + 4t = 10, so 9t + 1 = 10 and t = 1.
- 4Substitute t = 1 into the line: OF = (3, 1, 2) + (1, 2, −2) = (4, 3, 0), so the foot of perpendicular is F(4, 3, 0).
- 5Find the distance: |PF| = |t| |n| = 1 × √(1² + 2² + (−2)²) = √9 = 3 units.
- 6Check with the formula: p · n = (3)(1) + (1)(2) + (2)(−2) = 1, so the distance is |1 − 10| ÷ 3 = 3, which agrees.
- 7Use the reflection formula: OP′ = 2OF − OP = (8, 6, 0) − (3, 1, 2) = (5, 5, −2), so the image is P′(5, 5, −2).
- 8Check the midpoint: the midpoint of P(3, 1, 2) and P′(5, 5, −2) is (4, 3, 0) = F.
- 9Check the sides: OP′ · n = 5 + 10 + 4 = 19, and 19 − 10 = 9 while 1 − 10 = −9. The magnitudes are equal with opposite signs, so P and P′ are mirror images across π.
Watch out: The distance is |t| |n|, not |t|. The parameter t measures multiples of n, which is usually not a unit vector.
Common mistakes
Cosine instead of sine for a line–plane angle
Wrong: Using cos θ = |d · n| ÷ (|d| |n|) for the angle between a line and a plane.
Right: That formula gives the angle between the line and the normal, which is the complement of the angle you want. Use sin θ = |d · n| ÷ (|d| |n|). Using cosine gives 82.3° instead of the correct 7.7° in the line–plane example.
Declaring intersection from only two equations
Wrong: λ = 2, μ = 1 satisfy the first two component equations, so the lines intersect.
Right: You must substitute λ and μ into the third equation. If it fails, the lines are skew (given they are not parallel).
Mixing up direction vector and normal vector
Wrong: Line l is parallel to plane p because (1, 2, 2) is parallel to the plane's normal (2, 4, 4).
Right: A direction vector parallel to the normal means the line is perpendicular to the plane. For l ∥ p you need d · n = 0.
Questions students ask
How do I convert the vector equation of a line to cartesian form?
Write the three parametric equations x = a₁ + λd₁, y = a₂ + λd₂, z = a₃ + λd₃, then make λ the subject of each. Equate the results: (x − a₁) ÷ d₁ = (y − a₂) ÷ d₂ = (z − a₃) ÷ d₃. If a direction component is zero, that coordinate is constant and is written as a separate equation such as y = a₂; never divide by 0.
Why does the angle between a line and a plane use sine instead of cosine?
The dot product d · n gives the cosine of the angle between the line and the normal, which is 90° − θ. Since cos(90° − θ) = sin θ, the same computation gives the sine of the angle with the plane. Like with like (line–line, plane–plane) uses cosine; a mixed pair (direction against normal) uses sine.
How do I tell if two lines are parallel, intersecting or skew?
First check whether one direction vector is a scalar multiple of the other; if so, the lines are either parallel and distinct or the same line. If not, set a₁ + λd₁ = a₂ + μd₂ and solve two of the component equations for λ and μ. Then substitute into the third equation: if it is consistent the lines intersect, and if not they are skew.
Is d in r · n = d the distance from the origin to the plane?
Only when n is a unit vector. In general the distance from O to the plane is |d| ÷ |n|. For the plane x − y − 2z = −3, the distance is 3 ÷ √6 = √6 ÷ 2 ≈ 1.22, not 3.
In the full PDF
- •Labelled diagrams of a line through a point, a plane with its normal, a reflection seen edge-on, and the line–plane angle
- •Further worked examples: cartesian-to-vector conversion, a plane through three points, the foot of perpendicular to a line, and angles between two lines and between two planes
- •Fully worked intersection problems: two skew lines, a line meeting a plane, and the line of intersection of two planes, including the GC method (PlySmlt2 or rref)
- •The remaining exam red flags, including using a position vector as a direction and the wrong constant in r · n = d
- •An exam technique checklist and a summary of the five non-negotiables, with the sin-versus-cos angle table