| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| 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 C4 vectors question testing routine techniques: substituting a point into a line equation to find parameters, verifying a point lies on a line, using perpendicularity conditions (dot product = 0), and finding ratios along a line. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks |
|---|---|
| \(1 + 3\lambda = -5 \therefore \lambda = -2\) | M1 |
| \(p - \lambda = 9 \therefore p = 7\) | A1 |
| \(-5 + q\lambda = -9 \therefore q = 2\) | A1 |
| Answer | Marks |
|---|---|
| \(1 + 3\lambda = 25 \therefore \lambda = 8\) | M1 |
| when \(\lambda = 8\), \(\mathbf{r} = \begin{pmatrix} 1 \\ -5 \end{pmatrix} + 8 \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 25 \\ -1 \\ 11 \end{pmatrix}\) | |
| \(\therefore (25, -1, 11)\) lies on \(l\) | A1 |
| Answer | Marks |
|---|---|
| \(\overrightarrow{OC} = \begin{pmatrix} 1+3\lambda \\ 7-\lambda \\ -5+2\lambda \end{pmatrix} \therefore \begin{pmatrix} 7-\lambda \\ -5+2\lambda \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix} = 0\) | M1 |
| \(3 + 9\lambda - 7 + \lambda - 10 + 4\lambda = 0\) | |
| \(\lambda = 1 \therefore \overrightarrow{OC} = \begin{pmatrix} 4 \\ 6 \\ -3 \end{pmatrix}\), \(C(4, 6, -3)\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A: \lambda = -2, B: \lambda = 8, C: \lambda = 1 \therefore AC : CB = 3 : 7\) | M1 A1 | (10) |
## (i)
$1 + 3\lambda = -5 \therefore \lambda = -2$ | M1 |
$p - \lambda = 9 \therefore p = 7$ | A1 |
$-5 + q\lambda = -9 \therefore q = 2$ | A1 |
## (ii)
$1 + 3\lambda = 25 \therefore \lambda = 8$ | M1 |
when $\lambda = 8$, $\mathbf{r} = \begin{pmatrix} 1 \\ -5 \end{pmatrix} + 8 \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 25 \\ -1 \\ 11 \end{pmatrix}$ | |
$\therefore (25, -1, 11)$ lies on $l$ | A1 |
## (iii)
$\overrightarrow{OC} = \begin{pmatrix} 1+3\lambda \\ 7-\lambda \\ -5+2\lambda \end{pmatrix} \therefore \begin{pmatrix} 7-\lambda \\ -5+2\lambda \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix} = 0$ | M1 |
$3 + 9\lambda - 7 + \lambda - 10 + 4\lambda = 0$ | |
$\lambda = 1 \therefore \overrightarrow{OC} = \begin{pmatrix} 4 \\ 6 \\ -3 \end{pmatrix}$, $C(4, 6, -3)$ | M1 A1 |
## (iv)
$A: \lambda = -2, B: \lambda = 8, C: \lambda = 1 \therefore AC : CB = 3 : 7$ | M1 A1 | (10) |
---Relative to a fixed origin, $O$, the line $l$ has the equation
$$\mathbf{r} = \begin{pmatrix} 1 \\ p \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ q \end{pmatrix},$$
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=(\roman*)]
\item find the values of $p$ and $q$, [3]
\item show that the point $B$ with coordinates $(25, -1, 11)$ also lies on $l$. [2]
\end{enumerate}
The point $C$ lies on $l$ and is such that $OC$ is perpendicular to $l$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the coordinates of $C$. [3]
\item Find the ratio $AC : CB$ [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q6 [10]}}