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 carry all of 3D.
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 = d, and switch between them.
- Find a plane's normal from two directions lying in it.
WHAT YOU PROBABLY THINK
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:
Feed in λ and a point falls out; different λ, different point. Eliminating λ instead gives the cartesian form (x − a₁)/b₁ = (y − a₂)/b₂ = (z − a₃)/b₃, three fractions locked equal, which is the same line translated into coordinates.
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 is at right angles to n. Taking scalar products turns that sentence into one equation, not three, which disposes of the opening claim:
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.
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: 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: the plane is x − z = −1. Both B and C confirm: 2 − 3 = −1 and 0 − 1 = −1.
YOUR TURN
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.
THE EXAM BIT
- 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·(two directions) = 0 with a free choice of one component.
- Different anchors and scaled directions give equations that look different but 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
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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.
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 = d, and switch between them.
- Find a plane's normal from two directions lying in it.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.