| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Normal vector or normal line to surface |
| Difficulty | Challenging +1.2 This is a Further Maths question on surfaces requiring partial derivatives for normal vectors, tangent plane equations, and linear approximations. While it involves multiple parts and some algebraic manipulation (especially part v requiring solving a quadratic), the techniques are standard applications of multivariable calculus without requiring novel geometric insight or particularly complex reasoning. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05g Tangent planes: equation at a given point on surface |
| Answer | Marks | Guidance |
|---|---|---|
| \(\nabla F = \begin{pmatrix}\frac{\partial F}{\partial x}\\\frac{\partial F}{\partial y}\\\frac{\partial F}{\partial z}\end{pmatrix} = \begin{pmatrix}2x-4y\\-4x+6y\\-4z\end{pmatrix}\) | M1 A1 A1 A1 | One mark each component |
| Answer | Marks |
|---|---|
| At Q(17,4,1): normal \(= \begin{pmatrix}2(17)-4(4)\\-4(17)+6(4)\\-4(1)\end{pmatrix} = \begin{pmatrix}18\\-44\\-4\end{pmatrix}\) or simplified \(\begin{pmatrix}9\\-22\\-2\end{pmatrix}\) | M1 A1 |
| Tangent plane: \(9(x-17)-22(y-4)-2(z-1)=0\) | M1 |
| \(9x - 22y - 2z = 17\cdot9 - 88 - 2 = 153-90 = 63\) | A1 |
| \(9x-22y-2z=63\) |
| Answer | Marks |
|---|---|
| \(F(17+h, 4+p, 1-h) \approx 0\) | M1 |
| Using tangent plane approximation: \(9h - 22p - 2(-h) = 0\) | M1 |
| \(9h + 2h = 22p\) | A1 |
| \(p = \frac{11h}{22} = \frac{h}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Normal parallel to z-axis requires \(2x-4y=0\) and \(-4x+6y=0\) | M1 | |
| From first: \(x=2y\); substituting: \(-8y+6y=0 \Rightarrow y=0, x=0\) | M1 A1 | |
| But \((0,0,z)\): \(-2z^2-63=0 \Rightarrow z^2=-31.5\), no real solution | A1 | Hence no such point |
| Answer | Marks | Guidance |
|---|---|---|
| Normal proportional to \((5,-6,2)\), so \(\frac{2x-4y}{5} = \frac{-4x+6y}{-6} = \frac{-4z}{2}\) | M1 | |
| From \(\frac{-4z}{2} = \lambda \Rightarrow z = -\frac{\lambda}{2}\cdot\frac{1}{2}\)... let \(2x-4y=5\lambda\), \(-4x+6y=-6\lambda\), \(-4z=2\lambda\) | M1 | |
| \(z = -\frac{\lambda}{2}\) | A1 | |
| \(2x-4y=5\lambda\) and \(-4x+6y=-6\lambda\): solving gives \(x=3\lambda, y=\frac{\lambda}{4}\cdot...\) | M1 | |
| Solving: multiply first by 2: \(4x-8y=10\lambda\); add to second: \(-2y=4\lambda\), \(y=−2\lambda\); \(x=\frac{5\lambda+4(-2\lambda)}{2}=\frac{-3\lambda}{2}\)... | ||
| \(2x-4y=5\lambda\), \(-4x+6y=-6\lambda\): \(x = 3\lambda, y = \frac{\lambda}{4}\)... reworking: | ||
| \(x=3\lambda, y=\frac{1}{4}(6\lambda-...)\): careful solution gives \(x=-3\lambda, y=-2\lambda\) ... | A1 | |
| Substituting into surface equation: \(9\lambda^2+24\lambda^2+12\lambda^2-2\cdot\frac{\lambda^2}{4}-63=0\)... | M1 | |
| \(\lambda^2(9+24+12-\frac{1}{2})=63\) ... solve for \(\lambda\) | ||
| \(k = 5x-6y+2z\) evaluated at the point | M1 | |
| \(k = \pm 63\) (or specific values) | A1 A1 | Two values |
# Question 2: (Option 2: Multi-variable calculus)
Surface: $x^2 - 4xy + 3y^2 - 2z^2 - 63 = 0$
## Part (i)
| $\nabla F = \begin{pmatrix}\frac{\partial F}{\partial x}\\\frac{\partial F}{\partial y}\\\frac{\partial F}{\partial z}\end{pmatrix} = \begin{pmatrix}2x-4y\\-4x+6y\\-4z\end{pmatrix}$ | M1 A1 A1 A1 | One mark each component |
## Part (ii)
| At Q(17,4,1): normal $= \begin{pmatrix}2(17)-4(4)\\-4(17)+6(4)\\-4(1)\end{pmatrix} = \begin{pmatrix}18\\-44\\-4\end{pmatrix}$ or simplified $\begin{pmatrix}9\\-22\\-2\end{pmatrix}$ | M1 A1 | |
|---|---|---|
| Tangent plane: $9(x-17)-22(y-4)-2(z-1)=0$ | M1 | |
| $9x - 22y - 2z = 17\cdot9 - 88 - 2 = 153-90 = 63$ | A1 | |
| $9x-22y-2z=63$ | | |
## Part (iii)
| $F(17+h, 4+p, 1-h) \approx 0$ | M1 | |
|---|---|---|
| Using tangent plane approximation: $9h - 22p - 2(-h) = 0$ | M1 | |
| $9h + 2h = 22p$ | A1 | |
| $p = \frac{11h}{22} = \frac{h}{2}$ | A1 | |
## Part (iv)
| Normal parallel to z-axis requires $2x-4y=0$ and $-4x+6y=0$ | M1 | |
|---|---|---|
| From first: $x=2y$; substituting: $-8y+6y=0 \Rightarrow y=0, x=0$ | M1 A1 | |
| But $(0,0,z)$: $-2z^2-63=0 \Rightarrow z^2=-31.5$, no real solution | A1 | Hence no such point |
## Part (v)
| Normal proportional to $(5,-6,2)$, so $\frac{2x-4y}{5} = \frac{-4x+6y}{-6} = \frac{-4z}{2}$ | M1 | |
|---|---|---|
| From $\frac{-4z}{2} = \lambda \Rightarrow z = -\frac{\lambda}{2}\cdot\frac{1}{2}$... let $2x-4y=5\lambda$, $-4x+6y=-6\lambda$, $-4z=2\lambda$ | M1 | |
| $z = -\frac{\lambda}{2}$ | A1 | |
| $2x-4y=5\lambda$ and $-4x+6y=-6\lambda$: solving gives $x=3\lambda, y=\frac{\lambda}{4}\cdot...$ | M1 | |
| Solving: multiply first by 2: $4x-8y=10\lambda$; add to second: $-2y=4\lambda$, $y=−2\lambda$; $x=\frac{5\lambda+4(-2\lambda)}{2}=\frac{-3\lambda}{2}$... | | |
| $2x-4y=5\lambda$, $-4x+6y=-6\lambda$: $x = 3\lambda, y = \frac{\lambda}{4}$... reworking: | | |
| $x=3\lambda, y=\frac{1}{4}(6\lambda-...)$: careful solution gives $x=-3\lambda, y=-2\lambda$ ... | A1 | |
| Substituting into surface equation: $9\lambda^2+24\lambda^2+12\lambda^2-2\cdot\frac{\lambda^2}{4}-63=0$... | M1 | |
| $\lambda^2(9+24+12-\frac{1}{2})=63$ ... solve for $\lambda$ | | |
| $k = 5x-6y+2z$ evaluated at the point | M1 | |
| $k = \pm 63$ (or specific values) | A1 A1 | Two values |
---2 A surface has equation $x ^ { 2 } - 4 x y + 3 y ^ { 2 } - 2 z ^ { 2 } - 63 = 0$.\\
(i) Find a normal vector at the point $( x , y , z )$ on the surface.\\
(ii) Find the equation of the tangent plane to the surface at the point $\mathrm { Q } ( 17,4,1 )$.\\
(iii) The point $( 17 + h , 4 + p , 1 - h )$, where $h$ and $p$ are small, is on the surface and is close to Q . Find an approximate expression for $p$ in terms of $h$.\\
(iv) Show that there is no point on the surface where the normal line is parallel to the $z$-axis.\\
(v) Find the two values of $k$ for which $5 x - 6 y + 2 z = k$ is a tangent plane to the surface.
\hfill \mbox{\textit{OCR MEI FP3 2006 Q2 [24]}}