Maths › Further vectors › Lines and planes in three dimensions
Lines and planes in three dimensions
A line is a point plus a direction to slide along. A plane is a point plus a normal to stay perpendicular to. Two short equations describe both objects.
Builds on Vectors in three dimensions.
IN THIS TOPIC
- Write a line as r = a + λb and convert to cartesian form and back.
- Write a plane as r·n = d and as ax + by + cz = e, and switch between them.
- Find a plane's normal from two directions lying in it.
- Decide whether two lines meet, are parallel or are skew.
COMMON MISCONCEPTION
A plane needs three separate equations to describe, one for each coordinate.
Lines: a point and a direction
Every point of a line comes from one position vector a and some multiple of a direction vector b:
The booklet prints the cartesian form of a line but never this vector form, so learn it. Feed in λ and you get a point. Different λ, different point. Eliminating λ instead gives the cartesian form (x − a₁)/b₁ = (y − a₂)/b₂ = (z − a₃)/b₃, three fractions locked equal, the same line translated into coordinates, and that one is in the booklet under Vectors.
WORKED EXAMPLE
Line through two points
Find a vector equation of the line through P(3, 1, 2) and Q(5, 0, 4).
Direction: b = PQ = (2, −1, 2).
So r = (3, 1, 2) + λ(2, −1, 2), and the cartesian form is (x − 3)/2 = (y − 1)/(−1) = (z − 2)/2.
Either point works as the anchor. Swapping P for Q shifts λ by one and describes the identical line.
Planes: a point and a normal
A plane through point a, perpendicular to a normal vector n, contains exactly the points r for which r − a sits at right angles to n. Taking scalar products turns that sentence into one equation:
The scalar product form of a plane is not in the booklet, so learn it. Writing r = (x, y, z) and n = (n₁, n₂, n₃) expands it to the cartesian form n₁x + n₂y + n₃z = d. The normal's components sit in plain sight as the coefficients, which is the single most useful thing to notice about a plane in cartesian form. Watch the sign convention. The booklet writes a plane as n₁x + n₂y + n₃z + d = 0 with d = −a·n, so its d is the negative of the d used here. Decide which convention you are in before you quote anything.
WORKED EXAMPLE
The plane through three points
Find a cartesian equation of the plane through A(1, 0, 2), B(2, 1, 3) and C(0, 1, 1).
Two directions in the plane: AB = (1, 1, 1) and AC = (−1, 1, −1).
A normal n = (a, b, c) must satisfy n·AB = 0 and n·AC = 0, so a + b + c = 0 and −a + b − c = 0.
Adding gives b = 0, and then a = −c, so n = (1, 0, −1).
d = n·A = 1 − 2 = −1, and the plane is x − z = −1. Both B and C confirm it, since 2 − 3 = −1 and 0 − 1 = −1.
GUIDED PRACTICE
From scalar product form to cartesian
A plane has equation r·(1, 2, 2) = 5. Write the cartesian form, and decide whether the point (3, 1, 0) lies in it.
Show the working
Cartesian: x + 2y + 2z = 5.
Test the point. 3 + 2 + 0 = 5, so (3, 1, 0) lies in the plane.
The normal never moved. Converting forms is only unpacking the scalar product.
Do two lines meet?
In two dimensions any two lines either meet or are parallel. Three dimensions add a third option, skew, where the lines are not parallel and still never meet, like a motorway and the bridge above it. Test by setting the two position vectors equal. That gives three equations in λ and μ, so solve any two and then check the third. If it holds, the lines meet. If it fails and the directions are not multiples of each other, the lines are skew.
WORKED EXAMPLE
Two lines that do meet
Do r = (1, 0, 2) + λ(1, 2, −1) and r = (2, 5, −2) + μ(1, −1, 2) intersect?
Equate components: 1 + λ = 2 + μ, 2λ = 5 − μ, 2 − λ = −2 + 2μ.
From the first, μ = λ − 1. The second becomes 2λ = 6 − λ, so λ = 2 and μ = 1.
Check the third: 2 − 2 = 0 and −2 + 2 = 0. It holds, so the lines meet at (3, 4, 0).
The angle between them comes from the directions. (1, 2, −1)·(1, −1, 2) = −3, both lengths are √6, so cos θ = 3/6 and the acute angle is 60°.
Skipping the third check is the classic error here. Two equations in two unknowns will nearly always give you a λ and a μ, whether or not the lines actually meet, so the third component is the whole test.
ASSESSMENT FOCUS
- State direction vectors and normals explicitly. Most marks hang on identifying the right one.
- A cartesian line equation with a zero denominator means that coordinate is constant. Write it separately.
- To find a normal without a formula, solve n·(each direction) = 0 with a free choice of one component.
- Different anchors and scaled directions give equations that look different and earn the same marks.
CHECK YOURSELF
Does the point (7, −1, 5) lie on the line r = (1, 2, 3) + λ(2, −1, 1)?
Show a hint
Find λ from the x-coordinate, then test the other two.
Show the answer
From x: 1 + 2λ = 7, so λ = 3. Then y = 2 − 3 = −1, which matches, but z = 3 + 3 = 6 ≠ 5. One coordinate fails, so the point is not on the line.
Line: r = a + λb, one anchor point plus multiples of a direction.
Plane: r·n = d, every point whose displacement from the anchor is perpendicular to n.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the lines and planes in three dimensions questions page.
CHECK YOUR PROGRESS
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- Write a line as r = a + λb and convert to cartesian form and back.
- Write a plane as r·n = d and as ax + by + cz = e, and switch between them.
- Find a plane's normal from two directions lying in it.
- Decide whether two lines meet, are parallel or are skew.
Open the full revision checklist to see every objective in the course in one place.