A-LevelFurther MathematicsVectors9231/12 · May/June 2010 · Q710 Marks

The lines l₁ and l₂ have vector equations r = 4i – 2j + λ(2i + j – 4k) and r = 4i – 5j + 2k + μ(i – j – k) respectively.

(i)Show that l₁ and l₂ intersect.[3]
(ii)Find the perpendicular distance from the point P whose position vector is 3i – 5j + 6k to the plane containing l₁ and l₂.[3]
(iii)Find the perpendicular distance from P to l₁.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2010). The mark or mark type is shown in brackets.

[2] (i) Equate components of r: 4+2λ = 4+μ, -2+λ = -5-μ, -4λ = 2-μ. Solve any two equations to find λ = -1 and μ = -2.
[1] (i) Check consistency with the third equation: -4(-1) = 4, 2-(-2) = 4. Since 4=4, the values are consistent, so the lines intersect.
[1] (ii) Find normal vector to the plane using cross product of direction vectors: n = (2i + j – 4k) × (i – j – k) = (1(-1)-(-4)(-1))i - (2(-1)-(-4)(1))j + (2(-1)-1(1))k = (-1-4)i - (-2+4)j + (-2-1)k = -5i - 2j - 3k.
[1] (ii) Equation of the plane: r · n = a · n. Using point (4, -2, 0) from l1: (4i-2j) · (-5i-2j-3k) = -20+4 = -16. So -5x - 2y - 3z = -16, or 5x + 2y + 3z = 16.
[1] (ii) Perpendicular distance from P(3, -5, 6) to the plane: |(5(3) + 2(-5) + 3(6) - 16)| / √(5² + 2² + 3²) = |(15 - 10 + 18 - 16)| / √(25 + 4 + 9) = |7| / √38 = 7/√38 ≈ 1.14 (3 s.f.).
[1] (iii) Find vector AP, where A is a point on l₁ (4i - 2j): AP = P - A = (3i – 5j + 6k) - (4i – 2j) = -i - 3j + 6k. Direction vector of l₁ is d₁ = 2i + j – 4k. Cross product AP × d₁ = (-i - 3j + 6k) × (2i + j – 4k) = ((-3)(-4)-6(1))i - ((-1)(-4)-6(2))j + ((-1)(1)-(-3)(2))k = (12-6)i - (4-12)j + (-1+6)k = 6i + 8j + 5k.
[1] (iii) Set up formula for perpendicular distance: d = |AP × d₁| / |d₁|
[1] (iii) Calculate magnitude of numerator: |6i + 8j + 5k| = √(6² + 8² + 5²) = √(36 + 64 + 25) = √125.
[1] (iii) Calculate magnitude of denominator and final distance: |d₁| = √(2² + 1² + (-4)²) = √(4 + 1 + 16) = √21. So d = √125 / √21 = √(125/21) ≈ 2.44 (3 s.f.).