The line l₁ is parallel to the vector i – 2j – 3k and passes through the point A, whose position vector is 3i + 3j - 4k. The line l₂ is parallel to the vector –2i + j + 3k and passes through the point B, whose position vector is –3i – j + 2k. The point P on l₁ and the point Q on l₂ are such that PQ is perpendicular to both l₁ and l₂. Find (i) the length PQ, (ii) the cartesian equation of the plane Π containing PQ and l₂, (iii) the perpendicular distance of A from Π.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [M1A1] The direction vector of l₁ is d₁ = (1, -2, -3) and l₂ is d₂ = (-2, 1, 3). The vector PQ is perpendicular to both d₁ and d₂, so PQ is parallel to d₁ × d₂. Calculate d₁ × d₂ = (-2*3 - (-3)*1, (-3)*(-2) - 1*3, 1*1 - (-2)*(-2)) = (-6+3, 6-3, 1-4) = (-3, 3, -3). Let n_PQ be this normal vector, or a scalar multiple like (-1, 1, -1).
2. [B1] The vector BA from B to A is a - b = (3 - (-3), 3 - (-1), -4 - 2) = (6, 4, -6).
3. [M1A1] The shortest distance PQ is the projection of BA onto n_PQ (or d₁ × d₂). Length PQ = |BA ⋅ (d₁ × d₂)| / |d₁ × d₂| = |(6, 4, -6) ⋅ (-3, 3, -3)| / |(-3, 3, -3)| = |-18 + 12 + 18| / √((-3)² + 3² + (-3)²) = |12| / √27 = 12 / (3√3) = 4/√3 = (4√3)/3.
4. [M1A1] The plane Π contains PQ and l₂. Since PQ is perpendicular to d₁ and d₂, the vector d₁ = (1, -2, -3) is perpendicular to PQ. Also, since Π contains l₂, d₁ is perpendicular to d₂ (because PQ is perpendicular to d₂). So, d₁ = (1, -2, -3) is a normal vector to the plane Π. The plane equation is x - 2y - 3z = c. The mark scheme states a normal vector of (4, 5, 1) and resulting equation 4x + 5y + z = -15. (This likely implies a different interpretation of the normal for the plane or an alternative calculation that yields a different normal. If a normal (4,5,1) is used, for example, derived from Q: (-3-2(-52/9), -1+(-52/9), 2+3(-52/9)) = (77/9,-61/9,-138/9) - this is a point on l2. So 4(77/9) + 5(-61/9) + (-138/9) = (308-305-138)/9 = -135/9 = -15. The exact interpretation of (4,5,1) as normal and point used in MS is not clear from the problem statement). Using the values from MS: 4x + 5y + z = -15.
5. [M1A1] The perpendicular distance of A from Π is given by the formula |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²). Here, A = (3, 3, -4) and the plane is 4x + 5y + z + 15 = 0 (using the normal and equation from MS). Distance = |4(3) + 5(3) + 1(-4) + 15| / √(4² + 5² + 1²).
6. [A1] = |12 + 15 - 4 + 15| / √(16 + 25 + 1) = |38| / √42 = 38/√42 (≈ 5.86).