AS June 2019 Paper 1 Q13
13 Line \(l_1\) has Cartesian equation
\[x - 3 = \frac{2y + 2}{3} = 2 - z\](a) Write the equation of line \(l_1\) in the form\[\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\]
where \(\lambda\) is a parameter and \(\mathbf{a}\) and \(\mathbf{b}\) are vectors to be found. [2 marks]
(b) Line \(l_2\) passes through the points \(P(3, 2, 0)\) and \(Q(n, 5, n)\), where \(n\) is a constant.
(i) Show that the lines \(l_1\) and \(l_2\) are not perpendicular. [3 marks]
(ii) Explain briefly why lines \(l_1\) and \(l_2\) cannot be parallel. [2 marks]
(iii) Given that \(\theta\) is the acute angle between lines \(l_1\) and \(l_2\), show that\[\cos\theta = \frac{p}{\sqrt{34n^2 + qn + 306}}\]
where \(p\) and \(q\) are constants to be found. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Rewrites \(l_1\) in the general Cartesian form, or as a vector in terms of just one parameter. Or finds the position vector of a point on the line. Or finds a direction vector of \(l_1\) Or writes three equations expressing \(x\), \(y\) and \(z\) in terms of the parameter. | M1 | 3.1a |
| Writes \(l_1\) in a correct vector form. Accept \(\begin{pmatrix}x \\ y \\ z\end{pmatrix}\) in place of \(\mathbf{r}\) NMS can score 2/2 | A1 | 1.1b |
Typical solution
\[\frac{x - 3}{1} = \frac{y + 1}{1.5} = \frac{z - 2}{-1}\]\[\mathbf{r} = \begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 1.5 \\ -1\end{pmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| (i) Calculates the scalar product of their direction vectors of lines \(l_1\) and \(l_2\) Must not use a position vector as a direction vector. | M1 | 3.1a |
| Calculates a correct scalar product of a vector parallel to \(\begin{pmatrix}1 \\ 1.5 \\ -1\end{pmatrix}\) with a vector parallel to \(\begin{pmatrix}n - 3 \\ 3 \\ n\end{pmatrix}\) | A1 | 1.1b |
| Explains that, as the scalar product is not zero, then the lines are not perpendicular. Follow through their direction vectors if their scalar product is non-zero. Must follow M1. NMS scores 0/3 | E1F | 3.2a |
| (ii) Explains that two vectors are parallel if one is a multiple of the other. \(l_1\) and \(l_2\) need not be referred to explicitly. Possibly implied by \(\boldsymbol{c} = k\boldsymbol{d}\) seen where \(\boldsymbol{c}\) and \(\boldsymbol{d}\) are their direction vectors. | E1F | 2.4 |
| Demonstrates that, as \(n\) varies, the two direction vectors can never be a multiple of each other. | B1 | 3.1a |
| (iii) Uses the scalar product to form an equation in \(n\) and \(\cos\theta\) Follow through their direction vectors for lines \(l_1\) and \(l_2\) | M1 | 1.1a |
| Forms a correct equation in \(n\) and \(\cos\theta\) Or gives a correct expression for \(\cos\theta\) Accept a correct equation, or expression, for the supplementary angle. | A1 | 1.1b |
| Writes \(\cos\theta = \dfrac{3}{\sqrt{34n^2 - 102n + 306}}\) | R1 | 2.1 |
| (10 marks) |
Typical solution
(i)
\[\begin{pmatrix}1 \\ 1.5 \\ -1\end{pmatrix}\cdot\begin{pmatrix}n - 3 \\ 3 \\ n\end{pmatrix} = n - 3 + 4.5 - n = 1.5\]The scalar product is not zero \(\therefore\) lines \(l_1\) and \(l_2\) are not perpendicular.
(ii)
If \(\begin{pmatrix}2 \\ 3 \\ -2\end{pmatrix}\) and \(\begin{pmatrix}n - 3 \\ 3 \\ n\end{pmatrix}\) are parallel then
\[n - 3 = 2 \quad \text{and} \quad n = -2\]but \(n\) cannot be both 5 and \(-2\) \(\therefore\) \(l_1\) and \(l_2\) cannot be parallel
(iii)
\[\begin{pmatrix}2 \\ 3 \\ -2\end{pmatrix}\cdot\begin{pmatrix}n - 3 \\ 3 \\ n\end{pmatrix} = \sqrt{2^2 + 3^2 + 2^2} \times \sqrt{(n - 3)^2 + 3^2 + n^2} \times \cos\theta\]\[2(n - 3) + 9 - 2n = \sqrt{17} \times \sqrt{2n^2 - 6n + 18} \times \cos\theta\]\[\cos\theta = \frac{3}{\sqrt{34n^2 - 102n + 306}}\]