| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a straightforward vector lines question requiring standard techniques: equating components to find parameters, then verifying the intersection point. Part (ii) is trivial algebraic verification that one form is a scalar multiple of the other. Slightly easier than average due to being routine application of well-practiced methods with no conceptual challenges. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3+5t=1+2s\), \(2-3t=4-s\), \(-5+t=5+3s\) | M1 | attempt to solve any two simultaneously to obtain a value of \(s\) or \(t\) |
| \(t = -2\) and \(s = -4\) | A1 | |
| substitution of their \(s\) and \(t\) in other equation e.g. \(-7=-7\) or \(8=8\) | B1 | may be embedded e.g. \(-5 + -2 = 5 + 3\times -4\); B0 if any subsequent arithmetic errors |
| lines meet at \((-7, 8, -7)\) | A1 | allow in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-4\\2\\-6\end{pmatrix} = -2\begin{pmatrix}2\\-1\\3\end{pmatrix}\) seen | B1 | do not allow e.g. \(\begin{pmatrix}-4\\2\\-6\end{pmatrix} \div \begin{pmatrix}2\\-1\\3\end{pmatrix} = -2\) |
| common point identified and justified e.g. by substitution of correct value of \(s\) or \(u\), e.g. \(s=3\) or \(u=\frac{3}{2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| substitution of e.g. \(s = 3 - 2u\) | B1 | or e.g. \(u = \frac{3}{2}s - \frac{1}{2}\); or show one pair of equations consistent |
| and completion to \(\mathbf{r} = \begin{pmatrix}7\\1\\14\end{pmatrix} + u\begin{pmatrix}-4\\2\\-6\end{pmatrix}\) | B1 | and completion to \(\mathbf{r} = \begin{pmatrix}1\\4\\5\end{pmatrix} + u\begin{pmatrix}2\\-1\\3\end{pmatrix}\); show another pair consistent |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3+5t=1+2s$, $2-3t=4-s$, $-5+t=5+3s$ | M1 | attempt to solve any two simultaneously to obtain a value of $s$ or $t$ |
| $t = -2$ and $s = -4$ | A1 | |
| substitution of their $s$ and $t$ in other equation e.g. $-7=-7$ or $8=8$ | B1 | may be embedded e.g. $-5 + -2 = 5 + 3\times -4$; B0 if any subsequent arithmetic errors |
| lines meet at $(-7, 8, -7)$ | A1 | allow in vector form |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-4\\2\\-6\end{pmatrix} = -2\begin{pmatrix}2\\-1\\3\end{pmatrix}$ seen | B1 | do not allow e.g. $\begin{pmatrix}-4\\2\\-6\end{pmatrix} \div \begin{pmatrix}2\\-1\\3\end{pmatrix} = -2$ |
| common point identified and justified e.g. by substitution of correct value of $s$ or $u$, e.g. $s=3$ or $u=\frac{3}{2}$ | B1 | |
*Alternatively:*
| Answer | Marks | Guidance |
|--------|-------|----------|
| substitution of e.g. $s = 3 - 2u$ | B1 | or e.g. $u = \frac{3}{2}s - \frac{1}{2}$; or show one pair of equations consistent |
| and completion to $\mathbf{r} = \begin{pmatrix}7\\1\\14\end{pmatrix} + u\begin{pmatrix}-4\\2\\-6\end{pmatrix}$ | B1 | and completion to $\mathbf{r} = \begin{pmatrix}1\\4\\5\end{pmatrix} + u\begin{pmatrix}2\\-1\\3\end{pmatrix}$; show another pair consistent |
---5 The vector equations of two lines are as follows.
$$L : \mathbf { r } = \left( \begin{array} { l }
1 \\
4 \\
5
\end{array} \right) + s \left( \begin{array} { c }
2 \\
- 1 \\
3
\end{array} \right) \quad M : \mathbf { r } = \left( \begin{array} { c }
3 \\
2 \\
- 5
\end{array} \right) + t \left( \begin{array} { c }
5 \\
- 3 \\
1
\end{array} \right)$$
(i) Show that the lines $L$ and $M$ meet, and find the coordinates of the point of intersection.\\
(ii) Show that the line $L$ can also be represented by the equation $\mathbf { r } = \left( \begin{array} { c } 7 \\ 1 \\ 14 \end{array} \right) + u \left( \begin{array} { c } - 4 \\ 2 \\ - 6 \end{array} \right)$.
\hfill \mbox{\textit{OCR C4 2016 Q5 [6]}}