A2 June 2024 Q9
9.
The line \(l_2\) has equation \(\mathbf{r} = \begin{pmatrix}13\\ 5\\ 8\end{pmatrix} + \mu\begin{pmatrix}1\\ -2\\ 5\end{pmatrix}\)
where \(\lambda\) and \(\mu\) are scalar parameters.
The lines \(l_1\) and \(l_2\) intersect at the point \(P\).
Given that the plane \(\Pi\) contains both \(l_1\) and \(l_2\)
- pass through \((0, 0, 0)\)
- make an angle of \(60^\circ\) with the \(x\)-axis
- make an angle of \(45^\circ\) with the \(y\)-axis
| Scheme | Marks | AO |
|---|---|---|
| \(\left.\begin{aligned}2 + 3\lambda &= 13 + \mu\\ -3 + 4\lambda &= 5 - 2\mu\\ 1 - \lambda &= 8 + 5\mu\end{aligned}\right\}\) leading to \(\lambda = \ldots\) or \(\mu = \ldots\) (Note \(\lambda = 3, \mu = -2\)) | M1 | 3.1a |
| \(\begin{pmatrix}2\\ -3\\ 1\end{pmatrix} + 3\begin{pmatrix}3\\ 4\\ -1\end{pmatrix} = \begin{pmatrix}11\\ 9\\ -2\end{pmatrix}\) or \(\begin{pmatrix}13\\ 5\\ 8\end{pmatrix} - 2\begin{pmatrix}1\\ -2\\ 5\end{pmatrix} = \begin{pmatrix}11\\ 9\\ -2\end{pmatrix}\) | A1 | 1.1b |
| (2) |
Notes
Note: Accept any alternative vector forms of notation throughout.
M1: Forms and solves two equations to find the value of \(\lambda\) or \(\mu\).
A1: Correct point of intersection. Accept as coordinates or vector (oe).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 3 & 4 & -1\\ 1 & -2 & 5\end{vmatrix} = \mathbf{i}(20 - 2) - \mathbf{j}(15 + 1) + \mathbf{k}(-6 - 4)\) | M1 | 3.1a |
| \(\pm(18\mathbf{i} - 16\mathbf{j} - 10\mathbf{k})\) | A1 | 1.1b |
| e.g. \(\mathbf{r} \bullet \begin{pmatrix}18\\ -16\\ -10\end{pmatrix} = \begin{pmatrix}11\\ 9\\ -2\end{pmatrix} \bullet \begin{pmatrix}18\\ -16\\ -10\end{pmatrix} = \ldots\) | M1 | 1.1b |
| \(18x - 16y - 10z = 74\) o.e. \(\qquad k(9x - 8y - 5z = 37)\) | A1 | 2.5 |
| (4) |
Notes
M1: Finds the cross product of the direction vectors of the lines. If no method shown, two correct components implies the method.
A1: Correct normal vector.
M1: Dependent on having made some attempt at the cross product. Find the equation of a plane using \(\mathbf{r} \bullet (\text{their cross product}) = (\text{point on the plane}) \bullet (\text{their cross product}) = \ldots\) Condone minor miscopies of coordinates.
A1: Correct Cartesian equation of the plane. Accept any multiple of it.
(i)(b) Alt 1
| Scheme | Marks | AO |
|---|---|---|
| Plane is \(ax + by + cz = 1\), so \(\left\{\begin{aligned}2a - 3b + c &= 1\\ 13a + 5b + 8c &= 1\\ 11a + 9b - 2c &= 1\end{aligned}\right.\) | M1 A1 | 3.1a 1.1b |
| \(\left.\begin{aligned}2a - 3b + c &= 1\\ 13a + 5b + 8c &= 1\\ 11a + 9b - 2c &= 1\end{aligned}\right\} \Rightarrow \left.\begin{aligned}17a + c &= 4\\ 49a + 29c &= 8\end{aligned}\right\} \Rightarrow a = \ldots, b = \ldots, c = \ldots\) | M1 | 1.1b |
| \(\dfrac{9}{37}x - \dfrac{8}{37}y - \dfrac{5}{37}z = 1\) or \(9x - 8y - 5z = 37\) o.e. | A1 | 2.5 |
| (4) |
M1: Forms 3 equations in 3 unknowns using three points on the line (or may form four equations in four unknowns)
A1: All correct relevant equations.
M: Full process to solve the equations – which may be by calculator. Accept for any solutions appearing after setting up suitable equations.
A1: Correct Cartesian equation of the plane.
(i)(b) Alt 2
| Scheme | Marks | AO |
|---|---|---|
| Let normal vector be \(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) Then \((a\mathbf{i} + b\mathbf{j} + c\mathbf{k}) \bullet (3, 4, -1) = 0 = (a\mathbf{i} + b\mathbf{j} + c\mathbf{k}) \bullet (1, -2, 5) \Rightarrow\) \(\left.\begin{aligned}3a + 4b - c &= 0\\ a - 2b + 5c &= 0\end{aligned}\right\} \Rightarrow 10b - 16c = 0 \Rightarrow b = \dfrac{8}{5}c, a = -\dfrac{9}{5}c\) | M1 | 3.1a |
| \(\mathbf{n} = \pm k(9\mathbf{i} - 8\mathbf{j} - 5\mathbf{k})\) | A1 | 1.1b |
| Finds the Cartesian equation of the plane e.g. \(\mathbf{r} \bullet \mathbf{n} = (11\mathbf{i} + 9\mathbf{j} - 2\mathbf{k}) \bullet (9\mathbf{i} - 8\mathbf{j} - 5\mathbf{k}) = \ldots\) | M1 | 1.1b |
| \(9x - 8y - 5z = 37\) o.e. | A1 | 2.5 |
| (4) |
M1: Sets up the normal vector in terms of variables (may set one of them as 1 or another value) and takes the dot product with both directions to form and solve equations to find the normal.
A1: Correct normal vector.
M1: Dependent on having made some attempt at the normal vector. Finds the equation of a plane using \(\mathbf{r} \bullet (\text{their normal vector}) = (\text{point on the plane}) \bullet (\text{their normal vector}) = \ldots\) Condone minor miscopies of coordinates.
A1: Correct Cartesian equation of the plane. Accept any multiple of it.
| Scheme | Marks | AO |
|---|---|---|
| \(\cos^2 60^\circ + \cos^2 45^\circ + \cos^2\theta = 1\) leading to \(\cos\theta = \ldots\) \(\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{2}}{2}\right)^2 + \cos^2\theta = 1\) | M1 | 3.1a |
| \(\cos\theta = (\pm)\dfrac{1}{2}\) | A1 | 1.1b |
| \(\dfrac{x}{\cos 60^\circ} = \dfrac{y}{\cos 45^\circ} = \dfrac{z}{\text{their }\cos\theta}\) or \(\dfrac{x}{\left(\frac{1}{2}\right)} = \dfrac{y}{\left(\frac{\sqrt{2}}{2}\right)} = \dfrac{z}{\text{their }\cos\theta}\) | M1 | 1.1b |
| \(2x = \sqrt{2}y = 2z\) and \(2x = \sqrt{2}y = -2z\) o.e | A1 | 2.5 |
| (4) | ||
| (10 marks) |
Notes
M1: Uses \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\) to find the direction cosine to the \(z\)-axis. Allow with \(n\) or another letter in place of \(\cos\theta\)
A1: A correct answer for the direction cosine to the \(z\)-axis.
M1: Finds a Cartesian equation of one of the lines.
A1: Two correct Cartesian equations of the line, need not be simplified. Accept for both equation given, don’t be concerned if “and” or “or” is used.
(ii) Alt
| Scheme | Marks | AO |
|---|---|---|
| Let (unit) direction vector of line be \(\mathbf{d} = \begin{pmatrix}a\\ b\\ c\end{pmatrix}\) then \(\mathbf{d} \bullet \begin{pmatrix}1\\ 0\\ 0\end{pmatrix} = \|\mathbf{d}\|\cos 60^\circ = \dfrac{\|\mathbf{d}\|}{2}\) and \(\mathbf{d} \bullet \begin{pmatrix}0\\ 1\\ 0\end{pmatrix} = \|\mathbf{d}\|\cos 45^\circ = \dfrac{\|\mathbf{d}\|}{\sqrt{2}}\) | M1 | 3.1a |
| \(a = \dfrac{\|\mathbf{d}\|}{2}, b = \dfrac{\|\mathbf{d}\|}{\sqrt{2}}\) | A1 | 1.1b |
| \(\mathbf{d} = \|\mathbf{d}\|\left(\dfrac{1}{2}\mathbf{i} + \dfrac{1}{\sqrt{2}}\mathbf{j} + c\mathbf{k}\right) \Rightarrow c = \sqrt{1 - \dfrac{1}{4} - \dfrac{1}{2}} = \ldots\) \(\Rightarrow \mathbf{d} = \|\mathbf{d}\|\left(\dfrac{1}{2}\mathbf{i} + \dfrac{1}{\sqrt{2}}\mathbf{j} \pm \text{“}\dfrac{1}{2}\text{”}\mathbf{k}\right) \Rightarrow \dfrac{x}{1/2} = \dfrac{y}{1/\sqrt{2}} = \dfrac{z}{(\pm)1/2}\) | M1 | 1.1b |
| \(2x = \sqrt{2}y = 2z\) and \(2x = \sqrt{2}y = -2z\) o.e | A1 | 2.5 |
| (4) |
M1: Sets up the direction vector for the line in unknowns (use of unit vector is fine) and applies dot product with vectors in direction of \(x\) and \(y\) axes. Note, may quote the results directly as \(\cos 60^\circ = \dfrac{x}{\|\mathbf{d}\|}\) etc, for the projections on to the axes.
A1: Correct values for/expression in \(a\) and \(b\) (the \(x\) and \(y\) components of direction vector) or multiples of them.
M1: Full process to find the third ordinate for the direction vector, and proceeds to form the Cartesian equation of at least one of the lines.
A1: Two correct Cartesian equations of the line, need not be simplified.