| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a standard C4 vectors question testing perpendicularity (dot product = 0) and intersection of lines (equating components). Part (a) is routine, parts (b) and (c) require solving simultaneous equations in multiple variables but follow a well-practiced method with no novel insight needed. Slightly easier than average due to being a textbook-style multi-part question with clear signposting. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{vmatrix} 1 & 3 \\ 5 & b \end{vmatrix} = 0\) | M1 A1 | |
| \(\therefore 3 + 4a + 5b = 0\) | M1 A1 | |
| \(4 + s = -3 + 3t\) (1) | B1 | |
| \(1 + 4s = 1 + at\) (2) | M1 | |
| \(1 + 5s = -6 + bt\) (3) | M1 | |
| \((1) \Rightarrow s = 3t - 7\) | M1 | |
| sub. (2) \(\Rightarrow 1 + 4(3t - 7) = 1 + at\) | M1 A1 | |
| \(12t - 28 = at\), \(t(12 - a) = 28\), \(t = \frac{28}{12 - a}\) | M1 A1 | |
| sub. (3) \(\Rightarrow 1 + 5(3t - 7) = -6 + bt\) | A1 | |
| \(15t - 28 = bt\), \(t(15 - b) = 28\), \(t = \frac{28}{15 - b}\) | A1 | |
| \(\frac{28}{12 - a} = \frac{28}{15 - b}\), \(12 - a = 15 - b\), \(b = a + 3\) | M1 | |
| sub. (a) \(\Rightarrow 3 + 4a + 5(a + 3) = 0\), \(a = -2, b = 1\) | M1 A1 | |
| \(t = 2 \therefore r = \begin{pmatrix} -3 \\ -6 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\), \(\therefore (3, -3, -4)\) | M1 A1 | (12 marks) |
$\begin{vmatrix} 1 & 3 \\ 5 & b \end{vmatrix} = 0$ | M1 A1 |
$\therefore 3 + 4a + 5b = 0$ | M1 A1 |
$4 + s = -3 + 3t$ (1) | B1 |
$1 + 4s = 1 + at$ (2) | M1 |
$1 + 5s = -6 + bt$ (3) | M1 |
$(1) \Rightarrow s = 3t - 7$ | M1 |
sub. (2) $\Rightarrow 1 + 4(3t - 7) = 1 + at$ | M1 A1 |
$12t - 28 = at$, $t(12 - a) = 28$, $t = \frac{28}{12 - a}$ | M1 A1 |
sub. (3) $\Rightarrow 1 + 5(3t - 7) = -6 + bt$ | A1 |
$15t - 28 = bt$, $t(15 - b) = 28$, $t = \frac{28}{15 - b}$ | A1 |
$\frac{28}{12 - a} = \frac{28}{15 - b}$, $12 - a = 15 - b$, $b = a + 3$ | M1 |
sub. (a) $\Rightarrow 3 + 4a + 5(a + 3) = 0$, $a = -2, b = 1$ | M1 A1 |
$t = 2 \therefore r = \begin{pmatrix} -3 \\ -6 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$, $\therefore (3, -3, -4)$ | M1 A1 | (12 marks)
---Relative to a fixed origin, two lines have the equations
$$\mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} + s \begin{pmatrix} 1 \\ 4 \\ 5 \end{pmatrix}$$
and
$$\mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ -6 \end{pmatrix} + t \begin{pmatrix} 3 \\ a \\ b \end{pmatrix},$$
where $a$ and $b$ are constants and $s$ and $t$ are scalar parameters.
Given that the two lines are perpendicular,
\begin{enumerate}[label=(\alph*)]
\item find a linear relationship between $a$ and $b$. [2]
\end{enumerate}
Given also that the two lines intersect,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the values of $a$ and $b$, [8]
\item find the coordinates of the point where they intersect. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [12]}}