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 a plane, a line, or another line.
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.
- Apply the perpendicular distance formula from a point to a plane.
- Find the distance from a point to a line, and the shortest distance between two skew lines.
COMMON MISCONCEPTION
To find where a line meets a plane you must solve three simultaneous equations in three unknowns.
Piercing a plane
A line gives 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.
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, and which ratio to use
Angles come from the same two vectors, but the two situations do not use the same trig ratio. Between two planes, take the angle between their normals with cosine. Between a line and a plane, the direction b and the normal n give sin θ = |b·n|/(|b||n|), because the normal stands a right angle away from the plane itself. Use cosine there and you will hand in the complement of the right answer, which is worth nothing.
GUIDED PRACTICE
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), and 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. Cosine by mistake would have given 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:
The booklet prints this under Vectors, in the form with the plane written as n₁x + n₂y + n₃z + d = 0, so its numerator carries a plus and its d is the negative of the d used here. Same distance, opposite sign convention. Read the plane off the page carefully and the booklet does the rest. Substitute the point, subtract d, divide by the normal's length, and keep the modulus. A distance cannot be negative, and the sign only tells you 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.
Distances involving lines
There is no formula to quote here, only one idea used twice. Write down a general point on the line, form the joining vector, and force it perpendicular to every direction available. For a point to a line that is one condition and one unknown.
WORKED EXAMPLE
From a point to a line
Find the shortest distance from A(3, 0, 0) to the line r = λ(1, 2, 2).
A general point on the line is F = (λ, 2λ, 2λ), so AF = (λ − 3, 2λ, 2λ).
Perpendicularity: AF·(1, 2, 2) = λ − 3 + 4λ + 4λ = 9λ − 3 = 0, so λ = 1/3.
AF = (−8/3, 2/3, 2/3), whose length is 2√2.
Sanity check with Pythagoras. |OA| = 3, the foot sits 1 from the origin, and √(9 − 1) = 2√2.
Two skew lines need the same idea with two conditions. Take a general point on each, form the joining vector PQ, and demand PQ·b₁ = 0 and PQ·b₂ = 0. Solve the pair for λ and μ, then measure |PQ|. For r = (1, 0, 0) + λ(1, 1, 0) and r = (0, 0, 1) + μ(1, −1, 0), the two conditions give λ = −1/2 and μ = 1/2, so PQ = (0, 0, 1) and the shortest distance is 1. That answer is visible from the picture, since the lines live in horizontal planes one unit apart, which makes it a good first example. Harder lines only make the arithmetic longer; the two perpendicularity conditions never change.
ASSESSMENT FOCUS
- 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 you are doing.
- 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 'lying in it'.
- For shortest distances, show the perpendicularity conditions before solving. That is where the method marks are awarded.
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
Normals (1, 2, 2) and (1, 0, −1) have dot product 1 + 0 − 2 = −1, so |n₁·n₂| = 1. The lengths are 3 and √2, giving cos θ = 1/(3√2) and θ = 76.4°.
Line into plane: substitute the parametric coordinates and solve one equation in λ.
Line-plane angles use sine with the normal; plane-plane angles use cosine.
Shortest distances come from making the joining vector perpendicular to every direction in sight.
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 intersections, angles and distances questions page.
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.
- Apply the perpendicular distance formula from a point to a plane.
- Find the distance from a point to a line, and the shortest distance between two skew lines.
Open the full revision checklist to see every objective in the course in one place.