| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from origin to line |
| Difficulty | Standard +0.3 This is a standard multi-part vector line question requiring substitution to find constants, verification that a point lies on a line, using perpendicularity conditions (dot product = 0), and finding a ratio. All techniques are routine C4 material with straightforward algebraic manipulation and no novel problem-solving insight required. Slightly easier than average due to the step-by-step scaffolding. |
| Spec | 1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks |
|---|---|
| (a) \(1 + 3\lambda = -5 \therefore \lambda = -2\) | M1 |
| \(p - \lambda = 9 \therefore p = 7\) | A1 |
| \(-5 + 2\lambda = 11 \therefore q = 2\) | A1 |
| (b) \(1 + 3\lambda = 25 \therefore \lambda = 8\) | M1 |
| Answer | Marks |
|---|---|
| \(\therefore (25, -1, 11)\) lies on \(l\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore [(1 + 3\lambda)\mathbf{i} + (7 - 2\lambda)\mathbf{j} + (-5 + 2\lambda)\mathbf{k}] \cdot (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 0\) | M1 | |
| \(3 + 9\lambda - 7 + 2\lambda - 10 + 4\lambda = 0\) | A1 | |
| \(\lambda = 1 \therefore \overrightarrow{OC} = (4\mathbf{i} + 6\mathbf{j} - 3\mathbf{k})\), \(C(4, 6, -3)\) | M1 A1 | |
| (d) \(A: \lambda = -2\), \(B: \lambda = 8\), \(C: \lambda = 1\) \(\therefore AC : CB = 3 : 7\) | M1 A1 | (11 marks) |
**(a)** $1 + 3\lambda = -5 \therefore \lambda = -2$ | M1 |
$p - \lambda = 9 \therefore p = 7$ | A1 |
$-5 + 2\lambda = 11 \therefore q = 2$ | A1 |
**(b)** $1 + 3\lambda = 25 \therefore \lambda = 8$ | M1 |
When $\lambda = 8$, $\mathbf{r} = (1 + 3\mathbf{i} - 7\mathbf{j} - 5\mathbf{k}) + 8(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = (25\mathbf{i} - \mathbf{j} + 11\mathbf{k})$
$\therefore (25, -1, 11)$ lies on $l$ | A1 |
**(c)** $\overrightarrow{OC} = [(1 + 3\lambda)\mathbf{i} + (7 - 2\lambda)\mathbf{j} + (-5 + 2\lambda)\mathbf{k}]$
$\therefore [(1 + 3\lambda)\mathbf{i} + (7 - 2\lambda)\mathbf{j} + (-5 + 2\lambda)\mathbf{k}] \cdot (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 0$ | M1 |
$3 + 9\lambda - 7 + 2\lambda - 10 + 4\lambda = 0$ | A1 |
$\lambda = 1 \therefore \overrightarrow{OC} = (4\mathbf{i} + 6\mathbf{j} - 3\mathbf{k})$, $C(4, 6, -3)$ | M1 A1 |
**(d)** $A: \lambda = -2$, $B: \lambda = 8$, $C: \lambda = 1$ $\therefore AC : CB = 3 : 7$ | M1 A1 | (11 marks)
---3. Relative to a fixed origin, $O$, the line $l$ has the equation
$$\mathbf { r } = ( \mathbf { i } + p \mathbf { j } - 5 \mathbf { k } ) + \lambda ( 3 \mathbf { i } - \mathbf { j } + q \mathbf { k } ) ,$$
where $p$ and $q$ are constants and $\lambda$ is a scalar parameter.\\
Given that the point $A$ with coordinates $( - 5,9 , - 9 )$ lies on $l$,
\begin{enumerate}[label=(\alph*)]
\item find the values of $p$ and $q$,
\item show that the point $B$ with coordinates $( 25 , - 1,11 )$ also lies on $l$.
The point $C$ lies on $l$ and is such that $O C$ is perpendicular to $l$.
\item Find the coordinates of $C$.
\item Find the ratio $A C : C B$\\
3. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q3 [11]}}