| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| 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 FP3 vector geometry question requiring the perpendicular distance formula from a point to a line. While it involves 3D vectors and multiple steps (finding the connecting vector, taking cross product, computing magnitudes), it's a direct application of a well-practiced technique with no conceptual surprises. Slightly above average difficulty due to being Further Maths content and requiring careful arithmetic with the cross product. |
| Spec | 4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER \(c - a = \pm[1, 3, -2]\); \((c-a) \times [8, 3, -6]\); n = \(\pm[-12, 50, 9]\); \(d = \frac{ | \mathbf{n} | }{ |
| OR \(c-a = \pm[1,3,-2]\); \((c-a) \cdot [8,3,-6]\); \(\cos\theta = \pm\frac{109}{\sqrt{134}\sqrt{109}} = \pm\sqrt{\frac{109}{134}}\); \(d = \sqrt{134}\sin\theta\); \((\mathbf{d} = ) 5\) | B1; M1*; A1√; M1(dep*); A1; A1 | For vector joining lines; For attempt at scalar product of \(c-a\) and \([8,3,-6]\); For correct \(\cos\theta\) AEF. f.t. from incorrect \(c-a\); For using trigonometry for perpendicular distance; For correct expression for \(d\) in terms of \(\theta\); For correct distance CAO |
| OR \(c-a = \pm[1,3,-2]\); \((c-a) \cdot [8,3,-6]\); \(x = \frac{109}{\sqrt{109}} = \sqrt{109}\); \(d = \sqrt{134-109}\); \((\mathbf{d} = ) 5\) | B1; M1*; A1√; M1(dep*); A1; A1 | For vector joining lines; For attempt at scalar product of \(c-a\) and \([8,3,-6]\); For finding projection of \(c-a\) onto \([8,3,-6]\) f.t. from incorrect \(c-a\); For using Pythagoras for perpendicular distance; For correct expression for \(d\); For correct distance CAO |
| OR \(\mathbf{CP} = \pm[-11+8t, -3+3t, 2-6t]\); \(\mathbf{CP} \cdot [8,3,-6] = 0\); \(t = \pm1\) OR \(\mathbf{P} = (9, 5, -1)\); \(d = \sqrt{3^2+0^2+4^2}\); \((\mathbf{d} = ) 5\) | B1; M1*; A1√; M1(dep*); A1; A1 6 [6] | For finding a vector from C(12, 5, 3) to a point on the line; For using scalar product for perpendicularity with any vector involving \(t\); For correct point. f.t. from incorrect CP; For finding magnitude of CP; For correct expression for \(d\); For correct distance CAO |
**EITHER** $c - a = \pm[1, 3, -2]$; $(c-a) \times [8, 3, -6]$; **n** = $\pm[-12, 50, 9]$; $d = \frac{|\mathbf{n}|}{|[8,3,-6]|} = \frac{\sqrt{2725}}{\sqrt{109}} = (\mathbf{d} = ) 5$ | B1; M1*; A1√; M1(dep*); A1; A1 | For vector joining lines; For attempt at vector product of $c-a$ and $[8,3,-6]$; For obtaining **n**. f.t. from incorrect $c-a$; For dividing $|\mathbf{n}|$ by magnitude of $[8,3,-6]$; For either magnitude correct; For correct distance **CAO**
**OR** $c-a = \pm[1,3,-2]$; $(c-a) \cdot [8,3,-6]$; $\cos\theta = \pm\frac{109}{\sqrt{134}\sqrt{109}} = \pm\sqrt{\frac{109}{134}}$; $d = \sqrt{134}\sin\theta$; $(\mathbf{d} = ) 5$ | B1; M1*; A1√; M1(dep*); A1; A1 | For vector joining lines; For attempt at scalar product of $c-a$ and $[8,3,-6]$; For correct $\cos\theta$ **AEF**. f.t. from incorrect $c-a$; For using trigonometry for perpendicular distance; For correct expression for $d$ in terms of $\theta$; For correct distance **CAO**
**OR** $c-a = \pm[1,3,-2]$; $(c-a) \cdot [8,3,-6]$; $x = \frac{109}{\sqrt{109}} = \sqrt{109}$; $d = \sqrt{134-109}$; $(\mathbf{d} = ) 5$ | B1; M1*; A1√; M1(dep*); A1; A1 | For vector joining lines; For attempt at scalar product of $c-a$ and $[8,3,-6]$; For finding projection of $c-a$ onto $[8,3,-6]$ f.t. from incorrect $c-a$; For using Pythagoras for perpendicular distance; For correct expression for $d$; For correct distance **CAO**
**OR** $\mathbf{CP} = \pm[-11+8t, -3+3t, 2-6t]$; $\mathbf{CP} \cdot [8,3,-6] = 0$; $t = \pm1$ **OR** $\mathbf{P} = (9, 5, -1)$; $d = \sqrt{3^2+0^2+4^2}$; $(\mathbf{d} = ) 5$ | B1; M1*; A1√; M1(dep*); A1; A1 6 [6] | For finding a vector from C(12, 5, 3) to a point on the line; For using scalar product for perpendicularity with any vector involving $t$; For correct point. f.t. from incorrect **CP**; For finding magnitude of **CP**; For correct expression for $d$; For correct distance **CAO**
**SR** Obtain $\mathbf{CP} = [11, 3, -2] - [8, 3, -6] = \pm[3, 0, 4]$ B1; Verify $[3, 0, 4] \cdot [8, 3, -6] = 0$ M1*; $d = \sqrt{3^2+0^2+4^2} = 5$ M1(dep*) A1 A1 (maximum 5/6)
---Find the perpendicular distance from the point with position vector $12\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}$ to the line with equation $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 5\mathbf{k} + t(8\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})$. [6]
\hfill \mbox{\textit{OCR FP3 2006 Q3 [6]}}