| Exam Board | SPS |
|---|---|
| Module | SPS SM Mechanics (SPS SM Mechanics) |
| Year | 2022 |
| Session | February |
| Marks | 6 |
| Topic | Vectors Introduction & 2D |
| Type | Ratio division of line segment |
| Difficulty | Moderate -0.8 This is a straightforward 3D vectors question requiring only basic operations: finding magnitude using Pythagoras, using the parallelogram property that opposite sides are equal vectors, and applying ratio theorem for collinear points. All three parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it involves 3D vectors. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement |
Relative to a fixed origin $O$,
• the point $A$ has position vector $\mathbf{5i + 3j - 2k}$
• the point $B$ has position vector $\mathbf{7i + j + 2k}$
• the point $C$ has position vector $\mathbf{4i + 8j - 3k}$
\begin{enumerate}[label=(\alph*)]
\item Find $|\overrightarrow{AB}|$ giving your answer as a simplified surd.
[2]
\end{enumerate}
Given that $ABCD$ is a parallelogram,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the position vector of the point $D$.
[2]
\end{enumerate}
The point $E$ is positioned such that
• $ACE$ is a straight line
• $AC:CE = 2:1$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the coordinates of the point $E$.
[2]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Mechanics 2022 Q4 [6]}}