| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Perpendicular distance from point to line |
| Difficulty | Standard +0.3 This is a standard Further Maths vector question requiring finding the foot of perpendicular from a point to a line using dot product conditions, then a straightforward section formula calculation. While it involves multiple steps and Further Maths content, the techniques are routine and well-practiced for FP3 students with no novel problem-solving required. |
| Spec | 1.10d Vector operations: addition and scalar multiplication4.04a Line equations: 2D and 3D, cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1 (M, then distance) \(M = (1 + 2t, 1 + 3t, -1 + 2t)\) | B1 | For correct parametric form soi |
| \(\overrightarrow{AM} = (\pm)[2t - 6, 3t - 2, 2t - 8]\) | B1 FT | For correct vector. FT from \(M\) |
| \(\overrightarrow{AM} \text{ perp. } l \Rightarrow 2(2t-6) + 3(3t-2) + 2(2t-8) = 0\) | M1 | For using perpendicular condition |
| \(\Rightarrow t = 2, M = (5, 7, 3)\) | A1 A1 | For correct equation For correct coordinates |
| \(AM = \sqrt{2^2 + 4^2 + 4^2} = 6\) | M1 A1 | For using distance formula For correct distance |
| [7] | ||
| METHOD 2(a) (distance, then M) \((C = (1, 1, -1))\) \(\overrightarrow{AC} = \pm[6, 2, 8]\) | B1 | For correct vector |
| \(\mathbf{n} = \overrightarrow{AC} \times [2, 3, 2] = k[-20, 4, 14]\) | M1 | For finding \(\overrightarrow{AC} \times\) direction of \(l\) |
| \(d = \frac{\ | \mathbf{n}\ | }{[2,3,2]} = \frac{\sqrt{612}}{\sqrt{17}} = 6\) |
| A1 | For correct distance | |
| \(\overrightarrow{CM} = \sqrt{(6^2 + 2^2 + 8^2) - 6^2} = 2\sqrt{17}\) | M1 | For a correct method for finding position of \(M\) |
| \([2, 3, 2] = \sqrt{17} \Rightarrow t = 2, M = (5, 7, 3)\) | B1 | For \([2, 3, 2] = \sqrt{17}\) soi |
| A1 | ||
| [7] | ||
| METHOD 2(b) \((C = (1, 1, -1))\) \(\overrightarrow{AC} = \pm[6, 2, 8]\) | B1 | For correct vector |
| \(\cos\theta = \frac{\overrightarrow{AC} \cdot (2,3,2)}{[\overrightarrow{AC}]\ | (2,3,2)\ | }, \theta = 36.0(39...)\) or \(\sin\theta = \frac{153}{\sqrt{442}}\) |
| \(\ | \overrightarrow{AM}\ | = \ |
| \(M = (5, 7, 3)\) | M1,A1 | As above |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AM} = [-2, -4, -4]\) or \(\overrightarrow{MA} = [2, -4, 4]\) | M1 | For using \(A + k_1\overrightarrow{AM}\) or \(M + k_2\overrightarrow{MA}\) or ratio theorem or equivalent |
| \(\Rightarrow B = (7, 3, 7) + \frac{3}{4}(-2, -4, -4) = (7 - \frac{3}{2}, 3 + 3, 7 - 3)\) | M1 | For using \(B = (7, 3, 7) + \frac{3}{4}x'\)their \((-2,-4,-4)\) oe (or M1 for quadratic in parameter for line AM, followed by M1 for attempt to use correct value of parameter to find B) |
| A1 | For correct coordinates | |
| OR | ||
| \(B = (5, 7, 3) + \frac{1}{4}(7, 3, 7) = (\frac{15}{4}, \frac{7}{4}, \frac{21}{4} + \frac{3}{4}, \frac{9}{4} + \frac{7}{4})\) | M1 | |
| \(B = (\frac{11}{2}, 6, 4)\) | A1 | |
| [3] |
**(i)**
| METHOD 1 (M, then distance) $M = (1 + 2t, 1 + 3t, -1 + 2t)$ | B1 | For correct parametric form soi |
| $\overrightarrow{AM} = (\pm)[2t - 6, 3t - 2, 2t - 8]$ | B1 FT | For correct vector. FT from $M$ |
| $\overrightarrow{AM} \text{ perp. } l \Rightarrow 2(2t-6) + 3(3t-2) + 2(2t-8) = 0$ | M1 | For using perpendicular condition |
| $\Rightarrow t = 2, M = (5, 7, 3)$ | A1 A1 | For correct equation For correct coordinates |
| $AM = \sqrt{2^2 + 4^2 + 4^2} = 6$ | M1 A1 | For using distance formula For correct distance |
| | [7] | |
| METHOD 2(a) (distance, then M) $(C = (1, 1, -1))$ $\overrightarrow{AC} = \pm[6, 2, 8]$ | B1 | For correct vector |
| $\mathbf{n} = \overrightarrow{AC} \times [2, 3, 2] = k[-20, 4, 14]$ | M1 | For finding $\overrightarrow{AC} \times$ direction of $l$ |
| $d = \frac{\|\mathbf{n}\|}{[2,3,2]} = \frac{\sqrt{612}}{\sqrt{17}} = 6$ | A1 FT | For correct $\|\mathbf{n}\|$. FT from $\mathbf{n}$ |
| | A1 | For correct distance |
| $\overrightarrow{CM} = \sqrt{(6^2 + 2^2 + 8^2) - 6^2} = 2\sqrt{17}$ | M1 | For a correct method for finding position of $M$ |
| $[2, 3, 2] = \sqrt{17} \Rightarrow t = 2, M = (5, 7, 3)$ | B1 | For $[2, 3, 2] = \sqrt{17}$ soi |
| | A1 | |
| | [7] | |
| METHOD 2(b) $(C = (1, 1, -1))$ $\overrightarrow{AC} = \pm[6, 2, 8]$ | B1 | For correct vector |
| $\cos\theta = \frac{\overrightarrow{AC} \cdot (2,3,2)}{[\overrightarrow{AC}]\|(2,3,2)\|}, \theta = 36.0(39...)$ or $\sin\theta = \frac{153}{\sqrt{442}}$ | M1,A1 | |
| $\|\overrightarrow{AM}\| = \|\overrightarrow{AC}\|\sin\theta = 6$ | M1,A1 | |
| $M = (5, 7, 3)$ | M1,A1 | As above |
## Question 4 (continued)
**(ii)**
| $\overrightarrow{AM} = [-2, -4, -4]$ or $\overrightarrow{MA} = [2, -4, 4]$ | M1 | For using $A + k_1\overrightarrow{AM}$ or $M + k_2\overrightarrow{MA}$ or ratio theorem or equivalent |
| $\Rightarrow B = (7, 3, 7) + \frac{3}{4}(-2, -4, -4) = (7 - \frac{3}{2}, 3 + 3, 7 - 3)$ | M1 | For using $B = (7, 3, 7) + \frac{3}{4}x'$their $(-2,-4,-4)$ oe (or M1 for quadratic in parameter for line AM, followed by M1 for attempt to use correct value of parameter to find B) |
| | A1 | For correct coordinates |
| OR | | |
| $B = (5, 7, 3) + \frac{1}{4}(7, 3, 7) = (\frac{15}{4}, \frac{7}{4}, \frac{21}{4} + \frac{3}{4}, \frac{9}{4} + \frac{7}{4})$ | M1 | |
| $B = (\frac{11}{2}, 6, 4)$ | A1 | |
| | [3] | |4 The line $l$ has equations $\frac { x - 1 } { 2 } = \frac { y - 1 } { 3 } = \frac { z + 1 } { 2 }$ and the point $A$ is ( $7,3,7$ ). $M$ is the point where the perpendicular from $A$ meets $l$.\\
(i) Find, in either order, the coordinates of $M$ and the perpendicular distance from $A$ to $l$.\\
(ii) Find the coordinates of the point $B$ on $A M$ such that $\overrightarrow { A B } = 3 \overrightarrow { B M }$.
\hfill \mbox{\textit{OCR FP3 2012 Q4 [10]}}