| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Collinearity and ratio division |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: substituting coordinates into a line equation, using perpendicularity (dot product = 0), and checking collinearity with ratio division. All methods are routine C4 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10f Distance between points: using position vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
\begin{enumerate}
\item The point $A$, with coordinates $( 0 , a , b )$ lies on the line $l _ { 1 }$, which has equation
\end{enumerate}
$$\mathbf { r } = 6 \mathbf { i } + 19 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } )$$
(a) Find the values of $a$ and $b$.
The point $P$ lies on $l _ { 1 }$ and is such that $O P$ is perpendicular to $l _ { 1 }$, where $O$ is the origin.\\
(b) Find the position vector of point $P$.
Given that $B$ has coordinates $( 5,15,1 )$,\\
(c) show that the points $A , P$ and $B$ are collinear and find the ratio $A P : P B$.\\
\hfill \mbox{\textit{Edexcel C4 2006 Q5 [13]}}