| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| 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 straightforward multi-part vectors question requiring standard techniques: substituting points into a line equation to find parameters (part a), using perpendicularity condition with dot product (part b), and calculating distance (part c). All steps are routine applications of C4 vector methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\lambda = -4 \Rightarrow a = 18\); \(\mu = 1 \Rightarrow b = 9\) | M1, A1, A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\begin{pmatrix}8+\lambda\\12+\lambda\\14-\lambda\end{pmatrix} \cdot \begin{pmatrix}1\\1\\-1\end{pmatrix} = 0\) | M1 | |
| \(\therefore 8+\lambda+12+\lambda-14+\lambda = 0\) | A1 | |
| Solves to obtain \(\lambda = -2\) | dM1 | |
| Substitutes \(\lambda\) to give P at point \((6, 10, 16)\) (any form) | M1, A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(OP = \sqrt{36+100+256}\) | M1 | |
| \(= \sqrt{392} = 14\sqrt{2}\) | A1 cao | (2) [10] |
## Question 6:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\lambda = -4 \Rightarrow a = 18$; $\mu = 1 \Rightarrow b = 9$ | M1, A1, A1 | **(3)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}8+\lambda\\12+\lambda\\14-\lambda\end{pmatrix} \cdot \begin{pmatrix}1\\1\\-1\end{pmatrix} = 0$ | M1 | |
| $\therefore 8+\lambda+12+\lambda-14+\lambda = 0$ | A1 | |
| Solves to obtain $\lambda = -2$ | dM1 | |
| Substitutes $\lambda$ to give P at point $(6, 10, 16)$ (any form) | M1, A1 | **(5)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $OP = \sqrt{36+100+256}$ | M1 | |
| $= \sqrt{392} = 14\sqrt{2}$ | A1 cao | **(2) [10]** |
---6. The line $l _ { 1 }$ has vector equation
$$\mathbf { r } = 8 \mathbf { i } + 12 \mathbf { j } + 14 \mathbf { k } + \lambda ( \mathbf { i } + \mathbf { j } - \mathbf { k } ) ,$$
where $\lambda$ is a parameter.
The point $A$ has coordinates (4, 8, a), where $a$ is a constant. The point $B$ has coordinates ( $b , 13,13$ ), where $b$ is a constant. Points $A$ and $B$ lie on the line $l _ { 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$.
Given that the point $O$ is the origin, and that the point $P$ lies on $l _ { 1 }$ such that $O P$ is perpendicular to $l _ { 1 }$,
\item find the coordinates of $P$.
\item Hence find the distance $O P$, giving your answer as a simplified surd.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2006 Q6 [10]}}