| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard vectors question requiring perpendicularity (dot product = 0) and intersection (equating components). While it involves multiple parts and some algebraic manipulation, the techniques are routine for C4 level with no novel insight required. Slightly easier than average due to straightforward application of standard methods. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks |
|---|---|
| \(\begin{vmatrix} 1 & 3 \\ 4 & a \\ & b \end{vmatrix} = 0\) | M1 A1 |
| \(\therefore 3 + 4a + 5b = 0\) |
| Answer | Marks |
|---|---|
| \(4 + x = -3 + 3t\) ... (1) | |
| \(1 + 4x = 1 + at\) ... (2) | |
| \(1 + 5x = -6 + bt\) ... (3) | |
| From (1): \(x = 3t - 7\) | B1 M1 |
| Sub. (2): \(1 + 4(3t-7) = 1 + at\) | |
| \(12t - 28 = at\), \(t(12-a) = 28\), \(t = \frac{28}{12-a}\) | M1 A1 |
| Sub. (3): \(1 + 5(3t-7) = -6 + bt\) | |
| \(15t - 28 = bt\), \(t(15-b) = 28\), \(t = \frac{28}{15-b}\) | A1 |
| \(\frac{28}{12-a} = \frac{28}{15-b}\), \(12-a = 15-b\), \(b = a+3\) | M1 |
| Sub. (a): \(3 + 4a + 5(a+3) = 0\), \(a = -2\), \(b = 1\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(t = 2 \therefore \mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ -6 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ -4 \end{pmatrix}\) | M1 A1 |
| (12) |
**(i)**
$\begin{vmatrix} 1 & 3 \\ 4 & a \\ & b \end{vmatrix} = 0$ | M1 A1 |
$\therefore 3 + 4a + 5b = 0$ | |
**(ii)**
$4 + x = -3 + 3t$ ... (1) | |
$1 + 4x = 1 + at$ ... (2) | |
$1 + 5x = -6 + bt$ ... (3) | |
From (1): $x = 3t - 7$ | B1 M1 |
Sub. (2): $1 + 4(3t-7) = 1 + at$ | |
$12t - 28 = at$, $t(12-a) = 28$, $t = \frac{28}{12-a}$ | M1 A1 |
Sub. (3): $1 + 5(3t-7) = -6 + bt$ | |
$15t - 28 = bt$, $t(15-b) = 28$, $t = \frac{28}{15-b}$ | A1 |
$\frac{28}{12-a} = \frac{28}{15-b}$, $12-a = 15-b$, $b = a+3$ | M1 |
Sub. (a): $3 + 4a + 5(a+3) = 0$, $a = -2$, $b = 1$ | M1 A1 |
**(iii)**
$t = 2 \therefore \mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ -6 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ -4 \end{pmatrix}$ | M1 A1 |
| | **(12)** |7. Relative to a fixed origin, two lines have the equations\\
and
$$\begin{aligned}
& \mathbf { r } = \left( \begin{array} { c }
4 \\
1 \\
1
\end{array} \right) + s \left( \begin{array} { l }
1 \\
4 \\
5
\end{array} \right) \\
& \mathbf { r } = \left( \begin{array} { c }
- 3 \\
1 \\
- 6
\end{array} \right) + t \left( \begin{array} { l }
3 \\
a \\
b
\end{array} \right) ,
\end{aligned}$$
where $a$ and $b$ are constants and $s$ and $t$ are scalar parameters.\\
Given that the two lines are perpendicular,\\
(i) find a linear relationship between $a$ and $b$.
Given also that the two lines intersect,\\
(ii) find the values of $a$ and $b$,\\
(iii) find the coordinates of the point where they intersect.\\
\hfill \mbox{\textit{OCR C4 Q7 [12]}}