| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Equal length conditions |
| Difficulty | Challenging +1.2 Part (i) requires setting up equal length conditions |AC|=|AB| and algebraic manipulation to derive a linear constraint—straightforward vector magnitude work. Part (ii) involves minimizing triangle area subject to the constraint a-b+1=0, requiring calculus or geometric insight about perpendicular distance, making it a multi-step optimization problem that goes beyond routine exercises. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\vec{AC} = \begin{pmatrix}2-a\\4-b\\2\end{pmatrix},\ \vec{AB} = \begin{pmatrix}4-a\\2-b\\0\end{pmatrix}\) | M1 | AO 1.1 |
| \((4-a)^2 + (2-b)^2 = (2-a)^2 + (4-b)^2 + 4\) o.e. | M1 | AO 1.1a |
| \(16 - 8a + a^2 + 4 - 4b + b^2 = 4 - 4a + a^2 + 16 - 8b + b^2 + 4\) | M1 | AO 1.1 |
| \(4a - 4b + 4 = 0 \Rightarrow a - b + 1 = 0\) | A1 [4] | AO 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| D has position vector \(\begin{pmatrix}3\\3\\1\end{pmatrix}\) where D is midpoint of BC | B1 | Midpoint; OR if clearly minimising AC or AB - M1 for relevant vector using \(a\) and \(b\) (May be implied by second M1) |
| \(\overrightarrow{AD} = \begin{pmatrix}3-a\\2-a\\1\end{pmatrix}\) | M1 | Finding relevant vector in terms of \(a\) or \(b\) only |
| Area \(= \frac{1}{2}AD.BC = \frac{2\sqrt{3}\sqrt{(3-a)^2+(2-a)^2+1}}{2}\) | M1 | Expression for AD or \(AD^2\) (correct method but may have errors); May use area proportional to AD, AC or AB without calculation of expression for area |
| \(\sqrt{3}\sqrt{2\left((a-2.5)^2+0.75\right)}\) | M1 | Completion of square; Or differentiation of AD, \(AD^2\), AC, AB, \(AC^2\) or \(AB^2\) |
| \(a = 2.5\) for min | A1 | |
| Position vector \(\begin{pmatrix}2.5\\3.5\\0\end{pmatrix}\) | A1 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\vec{AC} = \begin{pmatrix}2-a\\4-b\\2\end{pmatrix},\ \vec{AB} = \begin{pmatrix}4-a\\2-b\\0\end{pmatrix}$ | M1 | AO 1.1 | Forming vectors for sides AB and AC; Implied by next M1 |
| $(4-a)^2 + (2-b)^2 = (2-a)^2 + (4-b)^2 + 4$ o.e. | M1 | AO 1.1a | Use of $AB = AC$ |
| $16 - 8a + a^2 + 4 - 4b + b^2 = 4 - 4a + a^2 + 16 - 8b + b^2 + 4$ | M1 | AO 1.1 | Expanding |
| $4a - 4b + 4 = 0 \Rightarrow a - b + 1 = 0$ | A1 [4] | AO 2.1 | AG Convincing completion |
## Question 10(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| D has position vector $\begin{pmatrix}3\\3\\1\end{pmatrix}$ where D is midpoint of BC | B1 | Midpoint; OR if clearly minimising AC or AB - M1 for relevant vector using $a$ and $b$ (May be implied by second M1) |
| $\overrightarrow{AD} = \begin{pmatrix}3-a\\2-a\\1\end{pmatrix}$ | M1 | Finding relevant vector in terms of $a$ or $b$ only |
| Area $= \frac{1}{2}AD.BC = \frac{2\sqrt{3}\sqrt{(3-a)^2+(2-a)^2+1}}{2}$ | M1 | Expression for AD or $AD^2$ (correct method but may have errors); May use area proportional to AD, AC or AB without calculation of expression for area |
| $\sqrt{3}\sqrt{2\left((a-2.5)^2+0.75\right)}$ | M1 | Completion of square; Or differentiation of AD, $AD^2$, AC, AB, $AC^2$ or $AB^2$ |
| $a = 2.5$ for min | A1 | |
| Position vector $\begin{pmatrix}2.5\\3.5\\0\end{pmatrix}$ | A1 | |
---10 Point A has position vector $\left( \begin{array} { l } a \\ b \\ 0 \end{array} \right)$ where $a$ and $b$ can vary, point B has position vector $\left( \begin{array} { l } 4 \\ 2 \\ 0 \end{array} \right)$ and point C has position vector $\left( \begin{array} { l } 2 \\ 4 \\ 2 \end{array} \right)$. ABC is an isosceles triangle with $\mathrm { AC } = \mathrm { AB }$.\\
(i) Show that $a - b + 1 = 0$.\\
(ii) Determine the position vector of A such that triangle ABC has minimum area.
Answer all the questions.\\
Section B (15 marks)
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q10 [10]}}