| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Easy -1.2 This is a straightforward question testing basic vector magnitude calculation using Pythagoras in 3D, followed by solving a simple inequality. Part (a) is pure recall/verification, and part (b) requires minimal problem-solving—just computing √(20+a²) > √38 and finding the smallest integer a satisfying a² > 18. |
| Spec | 1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\vert OA\vert =)\ \sqrt{5^2+3^2+2^2} = \sqrt{38}\) | B1* | Must see \(\sqrt{5^2+3^2+2^2}\) or e.g. \(\sqrt{25+9+4}\). Withhold for incorrect working such as \(\vert OA\vert = 5^2+3^2+2^2=38\Rightarrow\vert OA\vert=\sqrt{38}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\vert OB\vert = \sqrt{2^2+4^2+a^2}=\sqrt{20+a^2}\), so when \(a=5\), \(\vert OB\vert=\sqrt{20+25}=\sqrt{45}\) | M1 | Attempts \(\vert OB\vert\) in terms of \(a\), substitutes a positive integer for \(a\). May alternatively set up equation/inequality e.g. \(\sqrt{20+a^2}>\sqrt{38}\) and proceed to \(a^2>...\). |
| \(= 5\) | A1cso | Answer of 5 on its own with no incorrect working scores M1A1. Withhold if \(\vert OB\vert < \vert OA\vert\) is set at any point. |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\vert OA\vert =)\ \sqrt{5^2+3^2+2^2} = \sqrt{38}$ | B1* | Must see $\sqrt{5^2+3^2+2^2}$ or e.g. $\sqrt{25+9+4}$. Withhold for incorrect working such as $\vert OA\vert = 5^2+3^2+2^2=38\Rightarrow\vert OA\vert=\sqrt{38}$. |
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\vert OB\vert = \sqrt{2^2+4^2+a^2}=\sqrt{20+a^2}$, so when $a=5$, $\vert OB\vert=\sqrt{20+25}=\sqrt{45}$ | M1 | Attempts $\vert OB\vert$ in terms of $a$, substitutes a positive integer for $a$. May alternatively set up equation/inequality e.g. $\sqrt{20+a^2}>\sqrt{38}$ and proceed to $a^2>...$. |
| $= 5$ | A1cso | Answer of 5 on its own with no incorrect working scores M1A1. Withhold if $\vert OB\vert < \vert OA\vert$ is set at any point. |\begin{enumerate}
\item Relative to a fixed origin $O$
\end{enumerate}
\begin{itemize}
\item the point $A$ has position vector $5 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k }$
\item the point $B$ has position vector $2 \mathbf { i } + 4 \mathbf { j } + a \mathbf { k }$\\
where $a$ is a positive integer.\\
(a) Show that $| \overrightarrow { O A } | = \sqrt { 38 }$\\
(b) Find the smallest value of $a$ for which
\end{itemize}
$$| \overrightarrow { O B } | > | \overrightarrow { O A } |$$
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q3 [3]}}