A2 October 2021 Q7
7. With respect to a fixed origin \(O\), the line \(l\) has equation
\[(\mathbf{r} - (12\mathbf{i} + 16\mathbf{j} - 8\mathbf{k})) \times (9\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}) = \mathbf{0}\]The point \(A\) lies on \(l\) such that the direction cosines of \(\overrightarrow{OA}\) with respect to the \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) axes are \(\dfrac{3}{7}\), \(\beta\) and \(\gamma\).
Determine the coordinates of the point \(A\).
(7)
Way 1
| Scheme | Marks | AO |
|---|---|---|
| Position of \(A\) is given by \(\overrightarrow{OA} = \begin{pmatrix}12 + 9\lambda\\ 16 + 6\lambda\\ -8 + 2\lambda\end{pmatrix}\) | B1 | 3.1a |
| So have \(\dfrac{12 + 9\lambda}{\sqrt{(12 + 9\lambda)^2 + (16 + 6\lambda)^2 + (-8 + 2\lambda)^2}} = \dfrac{3}{7}\) | M1 | 1.1b |
| \(\Rightarrow 49\left(3(4 + 3\lambda)\right)^2 = 9\left((12 + 9\lambda)^2 + (16 + 6\lambda)^2 + (-8 + 2\lambda)^2\right)\) \(\Rightarrow 2880\lambda^2 + 7200\lambda + 2880 = 0\) or \(2\lambda^2 + 5\lambda + 2 = 0\) | M1 A1 | 3.1a 2.1 |
| \(\Rightarrow (2\lambda + 1)(\lambda + 2) = 0 \Rightarrow \lambda = \ldots\) | M1 | 1.1b |
| Substitutes a value of \(\lambda\) to find a position for A e.g. \(\overrightarrow{OA} = \begin{pmatrix}12 + 9\left(-\frac{1}{2}\right)\\ 16 + 6\left(-\frac{1}{2}\right)\\ -8 + 2\left(-\frac{1}{2}\right)\end{pmatrix} = \ldots\) | M1 | 1.1b |
| Coordinates of \(A\) are \(\left(\dfrac{15}{2}, 13, -9\right)\) only | A1 | 2.3 |
| (7) | ||
| (7 marks) |
Notes
Way 1
B1: Starts a correct procedure by parametrising the line correctly.
M1: Uses the direction cosine of \(\dfrac{3}{7}\) to form an equation in \(\lambda\)
M1: Realises need to square, to form quadratic in \(\lambda\) and gathers terms.
A1: Correct quadratic – three terms only or rearranged to complete square and solve, but need not have common factors all cancelled.
M1: Solves their three term quadratic, any valid method.
M1: Substitutes aa value for \(\lambda\) into the equation of the line to find a position for \(A\).
A1: Correct coordinates only
Way 2
| Scheme | Marks | AO |
|---|---|---|
| Direction of \(\overrightarrow{OA}\) is given by \(\mathbf{d} = \begin{pmatrix}\frac{3}{7}k\\ \beta k\\ \gamma k\end{pmatrix}\) or use of \(\left(\dfrac{3}{7}\right)^2 + \beta^2 + \gamma^2 = 1\) | B1 | 3.1a |
| \(\begin{pmatrix}\frac{3}{7}k - 12\\ \beta k - 16\\ \gamma k + 8\end{pmatrix} \times \begin{pmatrix}9\\ 6\\ 2\end{pmatrix} = \mathbf{0} \Rightarrow \left\{\begin{aligned}&6\left(\tfrac{3}{7}k - 12\right) - 9(\beta k - 16) = 0\\ &2\left(\tfrac{3}{7}k - 12\right) - 9(\gamma k + 8) = 0\\ &2(\beta k - 16) - 6(\gamma k + 8) = 0\end{aligned}\right.\) | M1 | 2.1 |
| \(\Rightarrow \beta k = \dfrac{2}{7}k + 8\) and \(\gamma k = \dfrac{1}{3}\left(\dfrac{2}{7}k - 32\right)\) \(\Rightarrow \dfrac{9k^2}{49} + \left(\dfrac{2}{7}k + 8\right)^2 + \dfrac{1}{9}\left(\dfrac{2}{7}k - 32\right)^2 = k^2 \Rightarrow 2k^2 - 7k - 490 = 0\) | M1 A1 | 3.1a 1.1b |
| \(\Rightarrow (2k - 35)(k + 14) = 0 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k \gt 0\) as direction cosine for first ordinate is positive, so need \(k = \dfrac{35}{2}\) hence \(\overrightarrow{OA} = \begin{pmatrix}\frac{3}{7} \times \frac{35}{2}\\ \frac{2}{7} \times \frac{35}{2} + 8\\ \frac{1}{3}\left(\frac{2}{7} \times \frac{35}{2} - 32\right)\end{pmatrix} = \ldots\) | M1 | 2.3 |
| Coordinates of \(A\) are \(\left(\dfrac{15}{2}, 13, -9\right)\) only | A1 | 1.1b |
| (7) |
B1: Starts correct procedure by using the direction cosines to parametrise \(\overrightarrow{OA}\) or attempting to use the Pythagorean property of the direction cosines.
M1: Uses their \(\overrightarrow{OA}\) as multiple of direction cosines in the line equation to produce simultaneous equations.
M1: Solves the system (no need to see check for consistency of third equation) to find \(\beta k\) and \(\gamma k\), or just \(\beta\) and \(\gamma\) in terms of \(k\) and proceeds to form a quadratic in \(k\) using the Pythagorean property of the direction cosines.
A1: A correct quadratic in \(k\) reduced to three terms etc.
M1: Solves their three term quadratic, any valid method.
M1: Substitutes aa value for \(\lambda\) into the equation of the line to find a position for \(A\).
A1: Correct coordinates only