Vectors, lines and planes in three dimensions
Use scalar and vector products with equations of lines and planes to solve intersections, angles and shortest-distance problems.
Section Paper 1 · Further Pure Mathematics 1 · Topics 1.1–1.7
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Convert between Cartesian, normal-vector and vector-parametric equations of planes
- Use scalar and vector products to find angles, areas and perpendicular directions
- Solve line-plane and plane-plane intersections
- Find distances between skew lines and equations of common perpendiculars
A normal vector connects the forms of a plane
The plane ax+by+cz=d has normal vector n=(a,b,c) and can be written r·n=d. A parametric plane r=a+λb+μc uses two non-parallel directions b and c; b×c is perpendicular to the plane.
A line r=p+td meets a plane when substituting it into the plane equation gives a consistent value of t. A zero coefficient of t means the line is parallel to the plane or lies in it; substitute a point to distinguish the cases. For two non-parallel planes, their intersection is a line whose direction is n₁×n₂; solve the two plane equations with a convenient free parameter to find a point on it.
The foot H from P to n·r=d lies on the normal line through P: H=P−((n·P−d)/|n|²)n. This projection formula also gives the point-to-plane distance |n·P−d|/|n|.
Products turn geometry into scalar conditions
The scalar product a·b=|a||b|cos θ gives angles and tests perpendicularity. The vector product a×b is perpendicular to both vectors and has magnitude |a||b|sin θ.
For skew lines r=p₁+sd₁ and r=p₂+td₂, a common-perpendicular direction is n=d₁×d₂. Their shortest distance is |(p₂−p₁)·n|/|n|. To find the common perpendicular itself, choose points on the two lines and solve for their connector to be perpendicular to both directions.
For a line and plane, the angle between the line direction and the plane normal is complementary to the line-plane angle. For two planes, the acute angle between their normals is the acute angle between the planes.
Pause and explain
Which habit best checks your work on vectors, lines and planes in three dimensions?
Worked example
Find where r=(1,0,0)+t(1,2,−1) meets the plane x+2y−z=6.
Show the worked solution
- Substitute x=1+t, y=2t and z=−t into the plane.
- (1+t)+4t−(−t)=6, so 1+6t=6 and t=5/6.
- Substitute t into the line to obtain the point.
Answer The intersection is (11/6, 5/3, −5/6).
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the angle between the planes x+2y+2z=4 and 2x−y+2z=1.
Give me a hint
The angle between planes is the acute angle between their normal vectors.
Compare my reasoning
- The normals are n₁=(1,2,2) and n₂=(2,−1,2).
- n₁·n₂=2−2+4=4; both magnitudes are 3.
- Thus cos θ=4/9 and θ=cos⁻¹(4/9).
cos⁻¹(4/9) ≈ 63.6°.
Look for these in your work
- Normals identified
- Dot product and magnitudes correct
- Acute plane angle stated
Find the shortest distance between the parallel planes 2x−y+2z=3 and 2x−y+2z=12.
Give me a hint
The normal has length 3; divide the difference in constants by that length.
Compare my reasoning
- Both planes share normal (2,−1,2), whose magnitude is 3.
- The perpendicular separation is |12−3|/3.
- The distance is positive and measured perpendicular to both planes.
3 units.
Look for these in your work
- Common normal used
- Constant difference absolute
- Distance is perpendicular
Lines L₁: r=(0,0,0)+s(1,0,0) and L₂: r=(0,1,1)+t(0,1,0) are skew. Find their shortest distance.
Give me a hint
Use a displacement between one point on each line and a unit vector along d₁×d₂.
Compare my reasoning
- The directions are d₁=(1,0,0), d₂=(0,1,0), so d₁×d₂=(0,0,1).
- A displacement from L₁'s point to L₂'s point is (0,1,1).
- Its component along the unit common-perpendicular direction has magnitude 1.
1 unit.
Look for these in your work
- Non-parallel common-perpendicular direction found
- Valid displacement chosen
- Perpendicular projection calculated
Common mistakes
- Treating a direction vector as a point
- Calling a parallel line an intersection without checking whether it lies in the plane
- Using the direction-normal angle instead of its complement for a line-plane angle
What can you explain now?
What is a normal vector to the plane 2x−y+3z=7?
Compare with the explanation
(2,−1,3)
The coefficients of x, y and z form a vector perpendicular to the plane.
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Make it stick
How do the scalar product, vector product and a plane's normal vector help solve 3D geometry problems?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.