Picture a pencil pushed through a sheet of paper: The meeting point of a line and a plane is on BOTH, so put the line's x, y, z into the plane's equation.
A line
- A moving point: r = a + λd.
- Its coordinates are x = a₁ + λd₁, y = a₂ + λd₂, z = a₃ + λd₃ — all written in terms of one parameter λ.
A plane
- All points (x, y, z) that satisfy one equation like n₁x + n₂y + n₃z = k.
- This is its Cartesian form, where (n₁, n₂, n₃) is the normal.
Finding the meeting point
Substitute
Put the line's x, y, z into the plane's equation.
Solve for λ
You get one equation in just λ. Solve it.
Feed λ back
Put λ back into the line to get the point.
IB-style question — find the intersection point
A line has equation r = (1, 0, 2) + λ(2, 1, −1). A plane has equation x + 2y + z = 9.
Find the coordinates of the point where the line meets the plane.
Step by step
- Write the line's coordinates in terms of λ — these are the (x, y, z) of any point on the line.
- The meeting point also obeys the plane's equation, so substitute these in.
- Expand and collect the λ-terms and the constants.
- Solve the single equation for λ.
- Put λ = 2 back into the line to get the actual point.
Final answer
The line meets the plane at (5, 2, 0).
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
When the parameter vanishes, something special is happening: After substituting, you normally get (number)·λ = number and solve for one λ. But sometimes the λ-terms cancel and you're left with just numbers.
False statement (e.g. 0 = 7)
- The line never satisfies the plane.
- It is parallel to the plane and misses it (no intersection).
True statement (e.g. 0 = 0)
- EVERY λ works.
- The line lies entirely in the plane (infinitely many common points).
Why the λ-terms cancel: The line's direction d is perpendicular to the plane's normal n (so d·n = 0). The line runs along the plane's surface direction; whether it sits in the plane or floats above it is decided by the constants.
IB-style question — show the line is parallel and misses
Show that the line r = (0, 1, 0) + λ(1, 1, 1) does not meet the plane x − 2y + z = 4.
Step by step
- The line's coordinates in terms of λ.
- Substitute into the plane's equation.
- Expand — watch the λ-terms.
- The λ-terms cancel (λ − 2λ + λ = 0), leaving an impossible statement.
- No λ can make this true, so there is no point on both.
Final answer
The statement −2 = 4 is impossible, so the line is parallel to the plane and never meets it.
IB-style question — show the line lies in the plane
Show that the line r = (2, 0, 1) + λ(1, 1, 1) lies in the plane x − 2y + z = 3.
Step by step
- The line's coordinates in terms of λ.
- Substitute into the plane's equation.
- Expand and collect.
- True for EVERY value of λ, so every point of the line is on the plane.
Final answer
The equation reduces to 3 = 3, true for all λ, so the whole line lies in the plane.