A2 June 2025 Paper 2 Q11
11 The line \(L\) has vector equation
\[\mathbf{r} = \begin{bmatrix} 5 \\ 2 \\ 11 \end{bmatrix} + \lambda\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\]The point \(A\) has coordinates \((1, -1, -6)\)
(a) Find the coordinates of the point on \(L\) which is closest to the point \(A\) [5 marks]
(b) The point \(B\) has coordinates \((3, -2, -1)\)
The point \(C\) has coordinates \((4, 0, 1)\)
The points \(A\), \(B\) and \(C\) all lie in the plane \(\Pi\)
Find a Cartesian equation of \(\Pi\) [4 marks]
(c) Find the coordinates of the point where the line \(L\) meets the plane \(\Pi\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Deduces a vector from \(A\) to a general point on the line. | M1 | 2.2a |
| Selects a suitable method, for example: Uses the scalar product Or Obtains an expression for \(|\overrightarrow{AP}|\) or \(|\overrightarrow{AP}|^2\) | M1 | 3.1a |
| Forms and solves an equation in \(\lambda\) | M1 | 1.1a |
| Substitutes their value of \(\lambda\) into the vector equation of \(L\) | M1 | 1.1a |
| Obtains \((-3, 6, -1)\) Condone position vector. | A1 | 1.1b |
| (5) |
Typical solution
At the closest point, \(\overrightarrow{AP} \bullet \underline{l} = 0\)
\[\overrightarrow{AP} = \begin{bmatrix} 4 + 2\lambda \\ 3 - \lambda \\ 17 + 3\lambda \end{bmatrix}\]\[\begin{bmatrix} 4 + 2\lambda \\ 3 - \lambda \\ 17 + 3\lambda \end{bmatrix} \bullet \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = 0\]\[8 + 4\lambda - 3 + \lambda + 51 + 9\lambda = 0\]\[56 + 14\lambda = 0\]\[\lambda = -4\]\[r = \begin{bmatrix} 5 \\ 2 \\ 11 \end{bmatrix} + (-4)\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -3 \\ 6 \\ -1 \end{bmatrix}\]The closest point is \((-3, 6, -1)\)
| Scheme | Marks | AO |
|---|---|---|
| Obtains two vectors in the plane. | M1 | 1.1a |
| Deduces a normal vector from their two vectors Or Obtains a vector equation of the plane in the form \(\mathbf{r} = \mathbf{a} + \lambda\overrightarrow{AB} + \mu\overrightarrow{BC}\) or similar. | M1 | 2.2a |
| Uses the position vector of \(A\), \(B\) or \(C\) to obtain the constant term Or Uses their vector equation to obtain and solve three simultaneous equations | M1 | 1.1a |
| Obtains a correct Cartesian equation. | A1 | 1.1b |
| (4) |
Typical solution
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{bmatrix} 2 \\ -1 \\ 5 \end{bmatrix} \times \begin{bmatrix} 3 \\ 1 \\ 7 \end{bmatrix} = \begin{bmatrix} -12 \\ 1 \\ 5 \end{bmatrix}\]\[-12 \times 4 + 1 \times 0 + 5 \times 1 = -43\]\[-12x + y + 5z = -43\]| Scheme | Marks | AO |
|---|---|---|
| Uses the equation of the line \(L\) and their equation of the plane \(\Pi\) to form an equation in \(\lambda\) | M1 | 3.1a |
| Solves equation in \(\lambda\) | M1 | 1.1a |
| Obtains \((13, -2, 23)\) Condone position vector. | A1 | 1.1b |
| (3) | ||
| (12 marks) |
Typical solution
\[-12x + y + 5z = -43\]\[-12(5 + 2\lambda) + (2 - \lambda) + 5(11 + 3\lambda) = -43\]\[-60 - 24\lambda + 2 - \lambda + 55 + 15\lambda + 43 = 0\]\[\lambda = 4\]The point is \((13, -2, 23)\)