| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question testing routine techniques: finding line of intersection via cross product of normals, checking parallelism via dot product, and verifying points lie on planes. All parts follow textbook methods with no novel insight required, though it requires more steps than typical A-level pure questions. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}1\\2\\1\end{pmatrix} \times \begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}7\\-2\\-3\end{pmatrix}\) | M1A1 | |
| Finds point on both planes | B1 | e.g. \((0,1,1)\); or \(\left(\frac{7}{3},\frac{1}{3},0\right)\) or \(\left(\frac{7}{2},0,-\frac{1}{2}\right)\) |
| \(\dfrac{x}{-7} = \dfrac{y-1}{2} = \dfrac{z-1}{3}\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x + 2y + z = 3\); \(2x + y + 4z = 5\) giving \(3x + 7z = 7\) and \(2x + 7y = 7\) | M1 A1 | Attempts at least 1 equation; 2 correct equations; or \(3y - 2z = 1\) |
| \(\dfrac{x}{-7} = \dfrac{y-1}{2} = \dfrac{z-1}{3}\) | M1A1 | oe of the form \(f(x)=g(y)=h(z)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-7\\2\\3\end{pmatrix}\cdot\begin{pmatrix}1\\5\\-1\end{pmatrix} = -7+10-3 = 0 \Rightarrow l \parallel \Pi_3\) | M1 | For scalar product, either shows method or gives answer of zero |
| A1 | For A1 must have working for scalar product | |
| \((0,1,1)\) is on line, but \(\begin{pmatrix}0\\1\\1\end{pmatrix}\cdot\begin{pmatrix}1\\5\\-1\end{pmatrix} = 4 \neq 1\), so not on plane | B1 | |
| ALT: \(x + 5y - z = 1\); \(7\lambda + 5(1-2\lambda)-(1-3\lambda)=1 \Rightarrow 4=1\) inconsistent, so \(l\) is parallel and not on plane | M1A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2 + 2\times0 + 1 = 3\) (so on \(\Pi_1\)) | Must show working for at least one plane | |
| \(\begin{pmatrix}2\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\5\\-1\end{pmatrix} = 1\) (so on \(\Pi_3\)) | B1 | Verify both |
| Line has equation \(\mathbf{r} = \begin{pmatrix}2\\0\\1\end{pmatrix} + \lambda\begin{pmatrix}-7\\2\\3\end{pmatrix}\) | M1 A1 | oe vector form; in Cartesian form M1 only; if cross product calculated incorrectly then M0A0 |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\2\\1\end{pmatrix} \times \begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}7\\-2\\-3\end{pmatrix}$ | M1A1 | |
| Finds point on both planes | B1 | e.g. $(0,1,1)$; or $\left(\frac{7}{3},\frac{1}{3},0\right)$ or $\left(\frac{7}{2},0,-\frac{1}{2}\right)$ |
| $\dfrac{x}{-7} = \dfrac{y-1}{2} = \dfrac{z-1}{3}$ | A1 | oe |
**ALT method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + 2y + z = 3$; $2x + y + 4z = 5$ giving $3x + 7z = 7$ and $2x + 7y = 7$ | M1 A1 | Attempts at least 1 equation; 2 correct equations; or $3y - 2z = 1$ |
| $\dfrac{x}{-7} = \dfrac{y-1}{2} = \dfrac{z-1}{3}$ | M1A1 | oe of the form $f(x)=g(y)=h(z)$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-7\\2\\3\end{pmatrix}\cdot\begin{pmatrix}1\\5\\-1\end{pmatrix} = -7+10-3 = 0 \Rightarrow l \parallel \Pi_3$ | M1 | For scalar product, either shows method or gives answer of zero |
| | A1 | For A1 must have working for scalar product |
| $(0,1,1)$ is on line, but $\begin{pmatrix}0\\1\\1\end{pmatrix}\cdot\begin{pmatrix}1\\5\\-1\end{pmatrix} = 4 \neq 1$, so not on plane | B1 | |
**ALT:** $x + 5y - z = 1$; $7\lambda + 5(1-2\lambda)-(1-3\lambda)=1 \Rightarrow 4=1$ inconsistent, so $l$ is parallel and not on plane | M1A1 A1 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2 + 2\times0 + 1 = 3$ (so on $\Pi_1$) | | Must show working for at least one plane |
| $\begin{pmatrix}2\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\5\\-1\end{pmatrix} = 1$ (so on $\Pi_3$) | B1 | Verify both |
| Line has equation $\mathbf{r} = \begin{pmatrix}2\\0\\1\end{pmatrix} + \lambda\begin{pmatrix}-7\\2\\3\end{pmatrix}$ | M1 A1 | oe vector form; in Cartesian form M1 only; if cross product calculated incorrectly then M0A0 |
---6 The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ have equations
$$\mathbf { r } \cdot \left( \begin{array} { l }
1 \\
2 \\
1
\end{array} \right) = 3 \text { and } \mathbf { r } \cdot \left( \begin{array} { l }
2 \\
1 \\
4
\end{array} \right) = 5$$
respectively. They intersect in the line $l$.\\
(i) Find cartesian equations of $l$.
The plane $\Pi _ { 3 }$ has equation $\mathbf { r } . \left( \begin{array} { c } 1 \\ 5 \\ - 1 \end{array} \right) = 1$.\\
(ii) Show that $\Pi _ { 3 }$ is parallel to $l$ but does not contain it.\\
(iii) Verify that $( 2,0,1 )$ lies on planes $\Pi _ { 1 }$ and $\Pi _ { 3 }$. Hence write down a vector equation of the line of intersection of these planes.\\
\hfill \mbox{\textit{OCR FP3 2016 Q6 [10]}}