The line \( l_1 \) passes through the point with position vector \( 8\mathbf{i} + 8\mathbf{j} - 7\mathbf{k} \) and is parallel to the vector \( 4\mathbf{i} + 3\mathbf{j} \). The line \( l_2 \) passes through the point with position vector \( 7\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} \) and is parallel to the vector \( 4\mathbf{i} - \mathbf{k} \). The point P on \( l_1 \) and the point Q on \( l_2 \) are such that PQ is perpendicular to both \( l_1 \) and \( l_2 \). In either order,

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

1. [M1A1] Uses vector product to find vector perpendicular to both lines. \( \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 0 \\ 4 & 0 & -1 \end{vmatrix} = -3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k} \)
2. [M1A1] Finds vecBA and its scalar product with unit perpendicular vector. vecBA = i + 10j - 11k perp.dist. = (|-3×1+4×10+12×11|)/(√(3² + 4² + 12²)) = 169/√169 = 13 (AG) (No penalty for sign errors made in n, which lead to correct result.)
3. [B1] Finds two parameter representation for PQ \( \mathbf{p} = \begin{pmatrix} 8+4\lambda \\ 8+3\lambda \\ -7 \end{pmatrix} \), \( \mathbf{q} = \begin{pmatrix} 7+4\mu \\ -2 \\ 4-\mu \end{pmatrix} \) \( \vec{PQ} = \begin{pmatrix} -1-4\lambda+4\mu \\ -10-3\lambda \\ 11-\mu \end{pmatrix} \)
4. [M1A1] Uses scalar product between PQ and direction vector of at least one line and equates to zero vecPQ · d₁ = 0 ⇒ 4(-1-4λ+4μ) + 3(-10-3λ) + 0(11-μ) = 0 vecPQ · d₂ = 0 ⇒ 4(-1-4λ+4μ) + 0(-10-3λ) -1(11-μ) = 0 16μ - 25λ = 34 17μ - 16λ = 15
5. [M1A1] Solves simultaneously. \( \mu = -1 \), \( \lambda = -2 \)
6. [A1] Obtains position vectors for P and Q \( \mathbf{p} = 2\mathbf{j} - 7\mathbf{k} \), \( \mathbf{q} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \)