AS June 2022 Paper 1 Q7
7 The lines \(l_1\) and \(l_2\) have equations
\[l_1 : \mathbf{r} = \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix} + \lambda\begin{bmatrix} 3 \\ -4 \\ 1 \end{bmatrix}\]\[l_2 : \mathbf{r} = \begin{bmatrix} -12 \\ a \\ -3 \end{bmatrix} + \mu\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}\](a) Show that the point \(P(-3, 9, -4)\) lies on \(l_1\) [2 marks]
(b) Show that \(l_1\) is perpendicular to \(l_2\) [2 marks]
(c) Given that the lines \(l_1\) and \(l_2\) intersect, calculate the value of the constant \(a\) [4 marks]
(d) Hence, find the coordinates of the point of intersection of \(l_1\) and \(l_2\) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Calculates a value of \(\lambda\) for point \(P\) Or writes a correct equation for \(l_1\) in Cartesian form (accept one error) and substitutes at least one of \(x = -3, \quad y = 9, \quad z = -4\) | M1 | 1.1a |
| Completes an argument to show that \(P\) lies on \(l_1\) Accept three correct calculations using \(\lambda = -2\) which lead to \(\begin{bmatrix} -3 \\ 9 \\ -4 \end{bmatrix}\) | R1 | 2.1 |
| (2) |
Typical solution
\[3 + 3\lambda = -3 \ \Rightarrow \ 3\lambda = -6 \ \Rightarrow \ \lambda = -2\]\[1 - 4\lambda = 9 \ \Rightarrow \ -4\lambda = 8 \ \Rightarrow \ \lambda = -2\]\[-2 + \lambda = -4 \ \Rightarrow \ \lambda = -2\]All three values of \(\lambda\) are the same so \(P\) lies on \(l_1\)
| Scheme | Marks | AO |
|---|---|---|
| Writes the scalar product of the two direction vectors. | M1 | 1.1a |
| Completes an argument to show that \(l_1\) and \(l_2\) are perpendicular. Accept, for two marks, \(3 \times 3 + (-4) \times 2 + 1 \times (-1) = 0\) so they are perpendicular Condone \(9 - 8 - 1 = 0\) with a reference to the scalar product. | R1 | 2.1 |
| (2) |
Typical solution
\[\begin{aligned}\begin{bmatrix} 3 \\ -4 \\ 1 \end{bmatrix}.\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix} &= 3 \times 3 + (-4) \times 2 + 1 \times (-1) \\ &= 9 - 8 - 1 \\ &= 0\end{aligned}\]\(\therefore\) \(l_1\) and \(l_2\) are perpendicular
| Scheme | Marks | AO |
|---|---|---|
| Selects a method to find the value of \(a\) eg equates the \(\boldsymbol{i}\) or \(\boldsymbol{k}\) component to form at least one equation in \(\lambda\) and \(\mu\) | M1 | 3.1a |
| Forms two correct equations in \(\lambda\) and \(\mu\) PI by a correct value of \(a\) PI by a correct value of \(\lambda\) or \(\mu\) | A1 | 1.1b |
| Calculates correct values of \(\lambda\) and \(\mu\) PI by a correct value of \(a\) | A1 | 1.1b |
| Obtains a correct value of \(a\) FT their \(\lambda\) and \(\mu\) | A1F | 1.1b |
| (4) |
Typical solution
\[3 + 3\lambda = -12 + 3\mu \ \Rightarrow \ \lambda = \mu - 5\]\[-2 + \lambda = -3 - \mu \ \Rightarrow \ \lambda = -1 - \mu\]\[\mu - 5 = -1 - \mu\]\[2\mu = 4\]\[\mu = 2\]\[\lambda = -1 - 2 = -3\]\[1 - 4\lambda = a + 2\mu\]\[1 - 4 \times (-3) = a + 2 \times 2\]\[a = 9\]| Scheme | Marks | AO |
|---|---|---|
| Obtains the correct coordinates of the point of intersection. Condone an answer of \(\begin{bmatrix} -6 \\ 13 \\ -5 \end{bmatrix}\) FT their \(\lambda\) or their \(a\) and \(\mu\) | B1F | 1.1b |
| (1) | ||
| (9 marks) |
Typical solution
\[\begin{bmatrix} 3 + (-3) \times 3 \\ 1 + (-3) \times (-4) \\ -2 + (-3) \times 1 \end{bmatrix} = \begin{bmatrix} -6 \\ 13 \\ -5 \end{bmatrix}\]Point of intersection \(= (-6, 13, -5)\)