Questions on intersections, angles and distances

Question 1

Find the point where the line r = (2, 0, 1) + λ(1, 2, 2) meets the plane x + y + z = 13.

Accepted Answer

Parametrise: x = 2 + λ, y = 2λ, z = 1 + 2λ. Substituting: (2 + λ) + 2λ + (1 + 2λ) = 13, so 5λ + 3 = 13 and λ = 2. The point is (4, 4, 5), and 4 + 4 + 5 = 13 confirms it. M1 for substituting the parametrisation into the plane, A1 for λ = 2, A1 for the point.

Question 2

State, with the correct trig function in each case, how to find the angle between a line and a plane, and between two planes.

Accepted Answer

Line and plane: sin θ = |b·n|/(|b||n|) with the line's direction and the plane's normal. Two planes: cos θ = |n₁·n₂|/(|n₁||n₂|) with the two normals. B1 for the sine formula, B1 for the cosine formula. Sine for the first because the normal stands at right angles to the plane.

Question 3

Find the point where the line r = (0, 1, 2) + λ(1, 2, 1) meets the plane 2x + y − z = 11.

Accepted Answer

Parametrise: x = λ, y = 1 + 2λ, z = 2 + λ. Substituting: 2λ + (1 + 2λ) − (2 + λ) = 11, so 3λ − 1 = 11 and λ = 4. The point is (4, 9, 6), and 2(4) + 9 − 6 = 11 confirms it. M1 for substituting the parametrisation into the plane, A1 for λ = 4, A1 for the point.

Question 4

In cos θ = |n₁·n₂|/(|n₁||n₂|) for the angle between two planes, explain why taking the modulus of n₁·n₂ guarantees an angle between 0° and 90°, even though the angle between the two normal vectors themselves could be obtuse.

Accepted Answer

Without the modulus, n₁·n₂ can be negative, which would make cos θ negative and θ obtuse: the angle between two normal vectors depends on which way each one happens to point, and either normal could just as well point the other way. B1. Taking the modulus removes the sign, so cos θ is never negative and θ never exceeds 90°, matching the convention that the angle between two planes is always taken to be the acute one. B1.

Question 5

Find the acute angle between the line with direction (1, 2, 2) and the plane x + y + z = 13, to one decimal place.

Accepted Answer

b·n = 1 + 2 + 2 = 5, |b| = 3 and |n| = √3. So sin θ = 5/(3√3) = 5√3/9, giving θ = 74.2°. M1 for b·n with both moduli, M1 for using sine, A1 for 5√3/9, A1 for the angle. Using cosine by habit would give the complement 15.8°, the angle to the normal instead of to the plane.

Question 6

Find the perpendicular distance from the point (1, 2, 3) to the plane 2x − y + 2z = 3.

Accepted Answer

Distance = |2 − 2 + 6 − 3|/√(4 + 1 + 4) = 3/3 = 1. M1 for substituting the point into the plane expression, M1 for dividing by the length of the normal, A1 for the distance.

Question 7

Find the acute angle between the planes x + z = 4 and y + z = 7.

Accepted Answer

Normals (1, 0, 1) and (0, 1, 1) have dot product 1 and lengths √2 and √2. cos θ = 1/2, so θ = 60°. M1 for both normals, M1 for the scalar product formula, A1 for cos θ = 1/2, A1 for the angle.

Question 8

Find the shortest distance from the point A(0, 9, 0) to the line r = λ(1, 2, 2), giving the answer as an exact surd.

Accepted Answer

There is no formula to quote here: write a general point and force perpendicularity. F = (λ, 2λ, 2λ) is a general point of the line, so AF = (λ, 2λ − 9, 2λ). M1 for the general point and the joining vector. The shortest join stands at right angles to the line: AF·(1, 2, 2) = λ + 2(2λ − 9) + 2(2λ) = 9λ − 18 = 0, so λ = 2. M1 for the perpendicularity condition, A1 for λ = 2. Then AF = (2, −5, 4), of length √(4 + 25 + 16) = √45 = 3√5. A1. Pythagoras confirms it: |OA| = 9, the foot F = (2, 4, 4) sits 6 from the origin, and √(81 − 36) = 3√5.

Question 9

Find the acute angle between the line with direction (1, −2, 2) and the plane 2x + y − 2z = 5, to one decimal place, explaining why the modulus in the formula is needed here.

Accepted Answer

b·n = (1)(2) + (−2)(1) + (2)(−2) = 2 − 2 − 4 = −4. M1 for the scalar product. |b| = 3 and |n| = 3, so sin θ = |−4|/9 = 4/9. M1 for the moduli, A1 for sin θ = 4/9. θ = 26.4°. A1. Here b·n itself is negative, so without the modulus the formula would give sin θ = −4/9, a negative sine for an angle that is supposed to lie between 0° and 90°: the modulus is what keeps the answer meaningful regardless of which way b or n happens to point. B1 for the explanation.

Question 10

Find the distance from the point (4, −1, 2) to the plane r·(2, 2, 1) = 6.

Accepted Answer

Substituting the point into the left side: 2(4) + 2(−1) + 1(2) = 8 − 2 + 2 = 8. M1 for the substitution. Distance = |8 − 6|/√(4 + 4 + 1) = 2/3. M1 for dividing by |n| and subtracting d, A1 for the value. The formula works exactly as it does for a cartesian plane, since r·n = d and n₁x + n₂y + n₃z = d are the same equation.

Question 11

A line has direction (3, 4, 0) and passes through the origin. Without finding a general point on the line, explain why the perpendicular distance from (0, 0, 5) to this line is exactly 5.

Accepted Answer

Every point of the line has the form (3λ, 4λ, 0), so the whole line lies in the plane z = 0. B1. The vector from the origin to (0, 0, 5) is (0, 0, 5), whose scalar product with the direction (3, 4, 0) is zero, so this vector is perpendicular to the line for every point on it, not just the origin. B1. Since the origin itself lies on the line, the perpendicular from (0, 0, 5) meets the line exactly at the origin. B1. The distance is therefore the length of (0, 0, 5), which is 5. B1. The usual method, forming a general point and imposing perpendicularity, would reach the same foot of the perpendicular but only after solving 25λ = 0.

Question 12

Show that the line r = (0, 0, 5) + λ(1, −1, 0) is parallel to the plane x + y + z = 1, determine whether it lies in the plane, and find the exact distance between them.

Accepted Answer

The direction is b = (1, −1, 0) and the normal is n = (1, 1, 1), so b·n = 1 − 1 + 0 = 0. The line runs at right angles to the normal, which puts it parallel to the plane, either inside it or clear of it. Testing the anchor point (0, 0, 5) gives 0 + 0 + 5 = 5 ≠ 1, so the line misses the plane. Because every point of the line is the same distance from the plane, that distance can be measured from the anchor: |0 + 0 + 5 − 1|/√(1 + 1 + 1) = 4/√3 = 4√3/3, about 2.31. M1 for b·n, A1 for the zero showing the line is parallel, M1 for testing the anchor point, A1 for concluding it does not lie in the plane, M1 for the distance formula, A1 for the exact distance. Substituting the line into the plane gives λ + (−λ) + 5 = 1, that is 5 = 1. The λ terms cancel and no solution survives, which is the algebraic signature of a line parallel to a plane it does not lie in.

Question 13

A student believes that finding where a line meets a plane means solving three simultaneous equations in three unknowns. Explain why one equation in one unknown is enough, and state what it means geometrically when that equation has no solution, and when every value of λ satisfies it.

Accepted Answer

The parametric line already hands over every one of its points as (x(λ), y(λ), z(λ)), so nothing remains to be solved for x, y and z separately: substituting those three expressions into the plane's cartesian equation leaves a single linear equation in λ, and its solution is the piercing point. B1 for the substitution collapsing the problem to one unknown. When the λ terms cancel and leave a false statement, no point of the line satisfies the plane: the line is parallel to the plane and clear of it. B1. When they cancel and leave a true statement, every point of the line satisfies the plane, so the line lies in the plane. B1.

Question 14

The line r = (0, 3, 1) + λ(2, 2, 1) meets the plane x + y + 2z = 17 at a point P. Find P, find the acute angle at which the line crosses the plane, find the perpendicular distance from (0, 3, 1) to the plane, and show that this distance equals the distance from (0, 3, 1) to P multiplied by the sine of the angle found.

Accepted Answer

Parametrise: x = 2λ, y = 3 + 2λ, z = 1 + λ. Substituting: 2λ + (3 + 2λ) + 2(1 + λ) = 17, so 6λ + 5 = 17 and λ = 2. M1 for substituting, A1 for λ = 2 and P = (4, 7, 3). For the angle, b·n = (2)(1) + (2)(1) + (1)(2) = 6, |b| = 3 and |n| = √6, so sin θ = 6/(3√6) = √6/3, giving θ = 54.7°. M1 for the scalar product and moduli, A1 for the angle. The distance from (0, 3, 1) to the plane is |0 + 3 + 2 − 17|/√6 = 12/√6 = 2√6. M1 for the distance formula. Along the line, (0, 3, 1) is a distance |λ||b| = 2 × 3 = 6 from P, and 6 × sin 54.7° = 6 × (√6/3) = 2√6, matching the perpendicular distance exactly: moving along the line from a point to where it meets the plane, the perpendicular drop is the length travelled times the sine of the angle to the plane, because that sine is defined as the ratio of the perpendicular component to the total length. A1 for the matching value and the reason.

Question 15

Let F be the foot of the perpendicular from a point A with position vector a to the plane n₁x + n₂y + n₃z = d. By writing F = a − tn for some scalar t, show that |AF| = |n₁a₁ + n₂a₂ + n₃a₃ − d|/|n|, and use this method, not the quoted formula, to find the distance from (5, 1, 2) to the plane 2x − y + 2z = 9.

Accepted Answer

Since AF is perpendicular to the plane, it is parallel to n, so F = a − tn for some scalar t. M1 for this setup. F lies in the plane, so n·F = d: n·a − t|n|² = d, giving t = (n·a − d)/|n|². M1 for substituting into the plane equation, A1 for t. Then AF = F − a = −tn, so |AF| = |t||n| = |n·a − d|/|n|² × |n| = |n₁a₁ + n₂a₂ + n₃a₃ − d|/|n|, as required. A1 for the general result. Applying it: n = (2, −1, 2), a = (5, 1, 2), so n·a = 10 − 1 + 4 = 13, |13 − 9| = 4, and |n| = 3. M1 for the substitution. The distance is 4/3. A1.