| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Common perpendicular to two skew lines |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple advanced techniques: finding the common perpendicular to skew lines (involving simultaneous equations with dot product conditions), cross products, and 3D geometry. While the individual steps are methodical, the multi-part structure, length, and need to coordinate several vector techniques places it well above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}\) | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2v}{dx^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\frac{d^2y}{dy^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \frac{2}{y^3}\left(\frac{dy}{dx}\right)^2\) | M1A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2}{y^3}\left(\frac{dy}{dx}\right)^2 - \frac{1}{y^2}\frac{d^2y}{dx^2} - \frac{2}{y^2}\frac{dy}{dx} + \frac{5}{y} = 17 + 6x - 5x^2 \sim \frac{d^2v}{dx^2} + 2\frac{dv}{dx} + 5v = 17 + 6x - 5x^2\) | A1 (4) | (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m^2 + 2m + 5 = 0 \Rightarrow m = -1 \pm 2i \Rightarrow CF: e^{-x}(A\cos 2x + B\sin 2x)\) | M1A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = px^2 + qx + r \Rightarrow v' = 2px + q \Rightarrow v'' = 2p\) | M1 | — |
| Equate coefficients: \(5p = -5\), \(4p + 5q = 6\), \(2p + 2q + 5r = 17\) | M1 | — |
| \(p = -1\), \(q = 2\), \(r = 3\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = e^{-x}(A\cos 2x + B\sin 2x) + 3 + 2x - x^2\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 0\): \(y = \frac{1}{2} \Rightarrow v = 2\) and \(\frac{dy}{dx} = -1 \Rightarrow \frac{dv}{dx} = 4\) | — | — |
| \(2 = A + 3 \Rightarrow A = -1\) | B1 | — |
| \(v' = e^{-x}(-2A\sin 2x + 2B\cos 2x) - e^{-x}(A\cos 2x + B\sin 2x)\) | M1 | — |
| \(4 = 2B + 1 + 2 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \left[e^{-x}\left(\frac{1}{2}\sin 2x - \cos 2x\right) + 3 + 2x - x^2\right]^{-1}\) | A1 (10), Total: 14 | — |
# Question 11(e):
**Substitution derivative:**
$\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}$ | B1 | —
**Second derivative:**
$\frac{d^2v}{dx^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\frac{d^2y}{dy^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \frac{2}{y^3}\left(\frac{dy}{dx}\right)^2$ | M1A1 | —
**Transformed equation:**
$\frac{2}{y^3}\left(\frac{dy}{dx}\right)^2 - \frac{1}{y^2}\frac{d^2y}{dx^2} - \frac{2}{y^2}\frac{dy}{dx} + \frac{5}{y} = 17 + 6x - 5x^2 \sim \frac{d^2v}{dx^2} + 2\frac{dv}{dx} + 5v = 17 + 6x - 5x^2$ | A1 (4) | (AG)
**Auxiliary equation and CF:**
$m^2 + 2m + 5 = 0 \Rightarrow m = -1 \pm 2i \Rightarrow CF: e^{-x}(A\cos 2x + B\sin 2x)$ | M1A1 | —
**Particular Integral:**
$v = px^2 + qx + r \Rightarrow v' = 2px + q \Rightarrow v'' = 2p$ | M1 | —
Equate coefficients: $5p = -5$, $4p + 5q = 6$, $2p + 2q + 5r = 17$ | M1 | —
$p = -1$, $q = 2$, $r = 3$ | A1 | —
**General solution:**
$v = e^{-x}(A\cos 2x + B\sin 2x) + 3 + 2x - x^2$ | A1 | —
**Applying initial conditions:**
When $x = 0$: $y = \frac{1}{2} \Rightarrow v = 2$ and $\frac{dy}{dx} = -1 \Rightarrow \frac{dv}{dx} = 4$ | — | —
$2 = A + 3 \Rightarrow A = -1$ | B1 | —
$v' = e^{-x}(-2A\sin 2x + 2B\cos 2x) - e^{-x}(A\cos 2x + B\sin 2x)$ | M1 | —
$4 = 2B + 1 + 2 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}$ | A1 | —
**Final answer:**
$y = \left[e^{-x}\left(\frac{1}{2}\sin 2x - \cos 2x\right) + 3 + 2x - x^2\right]^{-1}$ | A1 (10), **Total: 14** | —
---The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = 8 \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } )$ and $\mathbf { r } = 5 \mathbf { i } + 3 \mathbf { j } - 14 \mathbf { k } + \mu ( 2 \mathbf { j } - 3 \mathbf { k } )$ respectively. The point $P$ on $l _ { 1 }$ and the point $Q$ on $l _ { 2 }$ are such that $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$. Find the position vector of the point $P$ and the position vector of the point $Q$.
The points with position vectors $8 \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }$ and $5 \mathbf { i } + 3 \mathbf { j } - 14 \mathbf { k }$ are denoted by $A$ and $B$ respectively. Find\\
(i) $\overrightarrow { A P } \times \overrightarrow { A Q }$ and hence the area of the triangle $A P Q$,\\
(ii) the volume of the tetrahedron $A P Q B$. (You are given that the volume of a tetrahedron is $\frac { 1 } { 3 } \times$ area of base × perpendicular height.)
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\hfill \mbox{\textit{CAIE FP1 2015 Q11 OR}}