Edexcel AEA 2003 June — Question 1 5 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2003
SessionJune
Marks5
PaperDownload PDF ↗
TopicVectors Introduction & 2D
TypeGeometric properties using vectors
DifficultyChallenging +1.2 This is a well-structured geometric proof requiring students to find angles and side lengths in a triangle using vectors, then apply trigonometry. While it involves multiple steps (finding |OB|, |AB|, using dot product or geometry for angles, then tan formula), the path is relatively clear once the triangle is set up. It's harder than routine vector calculations but doesn't require the extended insight of the hardest AEA questions.
Spec1.05b Sine and cosine rules: including ambiguous case1.10b Vectors in 3D: i,j,k notation

1. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{25f0c7cc-0701-4836-931e-0eff5145e029-2_433_549_270_773}
\end{figure} The point \(A\) is a distance 1 unit from the fixed origin \(O\) .Its position vector is \(\mathbf { a } = \frac { 1 } { \sqrt { 2 } } ( \mathbf { i } + \mathbf { j } )\) . The point \(B\) has position vector \(\mathbf { a } + \mathbf { j }\) ,as shown in Figure 1. By considering \(\triangle O A B\) ,prove that \(\tan \frac { 3 \pi } { 8 } = 1 + \sqrt { } 2\) .

1.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
  \includegraphics[alt={},max width=\textwidth]{25f0c7cc-0701-4836-931e-0eff5145e029-2_433_549_270_773}
\end{center}
\end{figure}

The point $A$ is a distance 1 unit from the fixed origin $O$ .Its position vector is $\mathbf { a } = \frac { 1 } { \sqrt { 2 } } ( \mathbf { i } + \mathbf { j } )$ . The point $B$ has position vector $\mathbf { a } + \mathbf { j }$ ,as shown in Figure 1.

By considering $\triangle O A B$ ,prove that $\tan \frac { 3 \pi } { 8 } = 1 + \sqrt { } 2$ .

\hfill \mbox{\textit{Edexcel AEA 2003 Q1 [5]}}