AS June 2024 Paper 1 Q5
5 The line through points \(A(8, -7, -2)\) and \(B(11, -9, 0)\) is denoted by \(L_1\).
The line \(L_2\) passes through the origin, \(O\), and intersects \(L_1\) at the point \(C\). The lines \(L_1\) and \(L_2\) are perpendicular.
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} 11 \\ -9 \\ 0 \end{pmatrix} - \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) | M1 | 1.1 |
| \(\mathbf{r} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
M1: Subtracting the position vectors of \(A\) and \(B\) (in either order). Calculation or answer is enough for M1
Allow M1 for a row vector i.e. (3, -2, 2)
A1: Must be “\(\mathbf{r} =\)” (or “\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\)”)
Allow r or \(\mathbf{L_1}\) but not \(L_1\) (i.e. allow r not to be underlined as a vector, but if \(L_1\) used this must be indicated as a vector.
\(\mathbf{r} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix}\) or \(\begin{pmatrix} 11 \\ -9 \\ 0 \end{pmatrix} \pm \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\)
Note – could use other “starting points”
Must be column vectors here
| Scheme | Marks | AO |
|---|---|---|
| \(8 + 3\lambda = 26\) or \(-7 - 2\lambda = -19\) or \(-2 + 2\lambda = -14\) | M1 | 1.1 |
| (\(x\) gives \(\lambda = 6\) or \(y\) gives or \(\lambda = 6\)) [but] \(z\) gives \(\lambda = -6\) so \((26, -19, -14)\) does not lie on the line | A1FT | 2.2a |
| [2] |
Notes
M1: Writing down a correct equation for any component of their declared line.
Note – values of \(\lambda\) will depend on the “starting point” used in part (a) and different multiples of the direction vector.
A1FT: Finding a correct inconsistency and reaching correct conclusion. Not all values need be given but values given must be correct. FT their declared line.
Could see eg \(\lambda = 6 \Rightarrow\) (\(y = -19\) but) \(z = 10\) which is not \(-14\)
Do not need to see the word “inconsistent” or explicit comparison. Finding two different values of \(\lambda\) and then stating does not lie on line is enough
Alternate method
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 18 \\ -12 \\ -12 \end{pmatrix} = \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) | M1 |
| But \(\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) is not parallel to \(\begin{pmatrix} 18 \\ -12 \\ -12 \end{pmatrix}\) so the point does not lie on the line | A1FT |
M1: Rearranging vector equation
A1FT: Or finding \(\lambda\) inconsistency as before
A0 for “there is no value of \(\lambda\) that works” without justification
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) [for some particular value of \(\lambda\).] | B1FT | 3.1a |
| \(\left[\overrightarrow{OC} = \mu\begin{pmatrix} a \\ b \\ c \end{pmatrix}\right]\) and \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}.\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix} = 0\) | B1FT | 2.1 |
| \(3a - 2b + 2c = 0\) \(3(8 + 3\lambda) - 2(-7 - 2\lambda) + 2(-2 + 2\lambda) = 0\) | M1 | 1.1 |
| \(\Rightarrow \lambda = -2 \Rightarrow \overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) so the equation of \(L_2\) is \(\mathbf{r} = \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) | A1 | 2.2a |
| [4] |
Notes
B1FT: Can be embedded or implied. Can be same or different symbol as parameter in (a).
B1FT: Condition for perpendicularity stated.
Don’t have to see an equation for OC in terms of \(\mu\) here.
M1: Forming a correct dot product and using it with the other definition of \(\overrightarrow{OC}\) to form an equation in the parameter.
\(\mu\) may or may not be present here, but must be dealt with appropriately for the A mark.
A1: Condone use of the same parameter as \(L_1\).
Could see eg \(\mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\)
BOD lack of “\(\mathbf{r} =\)” (or “\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\)”) in this part if has already been penalised in part (a). If full marks awarded in part (a) then must have correct notation here for full marks.
Alternative Method
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) for some particular value of \(\lambda\). | B1FT |
| \(\overrightarrow{OC}.\mathbf{b}_{L_1} = 0\) | B1FT |
| \(\therefore \left(\begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\right).\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) \(= 24 + 14 - 4 + (9 + 4 + 4)\lambda = 0\) | M1 |
| \(\Rightarrow \lambda = -2 \Rightarrow \overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) so the equation of \(L_2\) is \(\mathbf{r} = \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) | A1 |
B1FT: Can be embedded or implied. Can be same or different symbol as parameter in (a).
B1FT: Condition for perpendicularity stated.
M1: Forming a correct dot product and using it to form an equation in the parameter.
\(\therefore \begin{pmatrix} 8 + 3\lambda \\ -7 - 2\lambda \\ -2 + 2\lambda \end{pmatrix}.\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix} = 24 + 9\lambda + 14 + 4\lambda - 4 + 4\lambda = 0\)
(Corrected from the printed mark scheme: the printed guidance has \(10 + 2\lambda\) as the third component and \(+20\) in the expansion; the third component of \(\overrightarrow{OC}\) is \(-2 + 2\lambda\), giving \(-4\).)
A1: Could see eg \(\mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\)
Condone presence of zero vector as first point
Condone use of the same parameter as \(L_1\).
BOD lack of “\(\mathbf{r} =\)” (or “\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\)”) in this part if has already been penalised in part (a). If full marks awarded in part (a) then must have correct notation here for full marks.
Alternative Method 2
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\) for some particular value of \(\lambda\). | B1FT |
| Distance \(|\overrightarrow{OC}|^2\) is given by: \(|\overrightarrow{OC}|^2 = (8 + 3\lambda)^2 + (-7 - 2\lambda)^2 + (-2 + 2\lambda)^2\) \(= 64 + 48\lambda + 9\lambda^2 + 49 + 28\lambda + 4\lambda^2 + 4 - 8\lambda + 4\lambda^2\) \(= 17\lambda^2 + 68\lambda + 117\) | M1 |
| Equating the derivative to 0 gives \(34\lambda + 68 = 0 \Rightarrow \lambda = -2\) | M1 |
| \(\Rightarrow \lambda = -2 \Rightarrow \overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) so the equation of \(L_2\) is \(\mathbf{r} = \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) | A1 |
B1FT: Can be embedded or implied. Can be same or different symbol as parameter in (a).
M1: Using the fact that minimum distance is the perpendicular distance.
A1: Condone use of the same parameter as \(L_1\).
Condone presence of zero vector as first point
Could see eg \(\mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\)
BOD lack of “\(\mathbf{r} =\)” (or “\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\)”) in this part if has already been penalised in part (a). If full marks awarded in part (a) then must have correct notation here for full marks.
| Scheme | Marks | AO |
|---|---|---|
| Because C lies on \(L_1\) and \(L_2\) and \(OC\) is perpendicular to \(L_1\), \(OC\) must be the shortest route from \(O\) to \(L_1\). | M1 | 3.1a |
| \(|\overrightarrow{OC}| = \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{49} = 7\) so shortest distance from \(O\) to \(L_1\) is 7 units. | A1 | 1.1 |
| [2] |
Notes
M1: Can be implied by sight of \(|\overrightarrow{OC}|\). Or attempt at evaluating \(|OC|\)
If just an evaluation seen need to see the vector identified as C in previous part (could be from equation). i.e. do not allow method mark for modulus of a vector which is not C.
A1: Follow through sign errors on coordinates of C