Maths › Further vectors › Intersections, angles and distances
Intersections, angles and distances
Where a line pierces a plane, the angle it makes on the way through, and how far a stray point sits from either: all scalar product work.
Builds on Lines and planes in three dimensions.
IN THIS TOPIC
- Substitute a parametric line into a plane's equation to find where they meet.
- Compute angles between a line and a plane, and between two planes, via normals.
- Apply the perpendicular distance formula from a point to a plane.
WHAT YOU PROBABLY THINK
To find where a line meets a plane you must solve three simultaneous equations in three unknowns.
Piercing a plane
A line hands you every one of its points as (x(λ), y(λ), z(λ)). Substitute those into the plane's cartesian equation and one linear equation in λ appears; its solution is the piercing point. One unknown, not three, which retires the opening claim.
WORKED EXAMPLE
A line meets a plane
Find where r = (1, 2, 0) + λ(1, −1, 2) meets the plane x + 2y + 2z = 11.
Parametrise: x = 1 + λ, y = 2 − λ, z = 2λ.
Substitute: (1 + λ) + 2(2 − λ) + 2(2λ) = 11, so 3λ + 5 = 11 and λ = 2.
The point is (3, 0, 4). Check: 3 + 0 + 8 = 11.
Angles come from the same two vectors. Between two planes, take the angle between their normals. Between a line and a plane, the direction b and normal n give sin θ = |b·n|/(|b||n|): sine, not cosine, because the normal stands a right angle away from the plane itself.
YOUR TURN
The angle of entry
Find the angle between the line with direction b = (1, −1, 2) and the plane x + 2y + 2z = 11.
Show the working
n = (1, 2, 2); b·n = 1 − 2 + 4 = 3.
|b| = √6 and |n| = 3, so sin θ = 3/(3√6) = 1/√6.
θ = 24.1° to one decimal place. Had cosine been used by mistake, the answer would be the complement, 65.9°, a classic dropped mark.
How far from a plane?
The perpendicular distance from a point to the plane n₁x + n₂y + n₃z = d comes from projecting onto the unit normal:
Substitute the point, subtract d, divide by the normal's length, and keep the modulus: distance cannot be negative, and the sign only says which side of the plane the point sits on.
WORKED EXAMPLE
A short drop
Find the distance from (2, 3, 1) to the plane x + 2y + 2z = 11.
Numerator: |2 + 6 + 2 − 11| = 1.
Denominator: √(1 + 4 + 4) = 3.
Distance = 1/3. The origin, for comparison, sits |0 − 11|/3 = 11/3 away on the other side.
THE EXAM BIT
- Substituting the parametric line into the plane is the intended route; say what λ you found.
- Line and plane use sine with the normal; two planes use cosine of their normals. State which.
- Quote the distance formula with the modulus in place; a negative distance loses the mark.
- If b·n = 0 the line is parallel to the plane: test a point to separate 'parallel' from 'inside'.
CHECK YOURSELF
Find the acute angle between the planes x + 2y + 2z = 11 and x − z = −1, to one decimal place.
Show a hint
cos θ = |n₁·n₂|/(|n₁||n₂|) with the two normals.
Show the answer
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Line into plane: substitute the parametric coordinates, solve one equation in λ.
Distances project onto the unit normal; line-plane angles use sine, plane-plane use cosine.
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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- Substitute a parametric line into a plane's equation to find where they meet.
- Compute angles between a line and a plane, and between two planes, via normals.
- Apply the perpendicular distance formula from a point to a plane.
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.