Year: 2019
Paper: 2
Question Number: 7
Course: LFM Pure and Mechanics
Section: Vectors
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, respectively. Each of these vectors is a unit vector (so $\mathbf{a} \cdot \mathbf{a} = 1$, for example) and
$$\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}.$$
Show that $\mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$. What can be said about the triangle ABC? You should justify your answer.
\item The four distinct points $A_i$ ($i = 1, 2, 3, 4$) have unit position vectors $\mathbf{a}_i$ and
$$\sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}.$$
Show that $\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4$.
\begin{enumerate}
\item Given that the four points lie in a plane, determine the shape of the quadrilateral with vertices $A_1$, $A_2$, $A_3$ and $A_4$.
\item Given instead that the four points are the vertices of a regular tetrahedron, find the length of the sides of this tetrahedron.
\end{enumerate}
\end{questionparts}\begin{questionparts}
\item Given $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$, we can form the following results:
\begin{align*}
&& \begin{cases} \mathbf{a} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\
\mathbf{b} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\
\mathbf{c} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \end{cases} \\
\Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= 0 \\
\mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= 0 \\
\mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} &= 0 \\ \end{cases} \\
\Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\
\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\
\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \end{cases} \\
\Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\
\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\
\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\
\mathbf{a} \cdot \mathbf{b} +\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -\frac12 \\ \end{cases} \\
\Rightarrow && \begin{cases}
\mathbf{a} \cdot \mathbf{b} = -\frac12 \\
\mathbf{a} \cdot \mathbf{c} = -\frac12 \\
\mathbf{b} \cdot \mathbf{c} = -\frac12 \\
\end{cases}
\end{align*}
The triangle must be equilateral since the angles between each vertex are the same.
\item We have $\displaystyle \sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}$ so $\displaystyle \mathbf{a}_i \cdot \sum_{i=1}^{4} \mathbf{a}_i = 0$ or for each $i$, $\displaystyle \sum_{j \neq i} \mathbf{a}_i \cdot \mathbf{a}_j = -1$.
\begin{align*}
&& \begin{cases}
\mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_1 \cdot \mathbf{a}_4 = -1 \\
\mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_4 = -1 \\
\mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\
\mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\
\end{cases} \\
&& \text{adding the first two, subtracting the last two} \\
\Rightarrow && \begin{cases}
\mathbf{a}_1 \cdot \mathbf{a}_2 +\cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} = -1 \\
\mathbf{a}_1 \cdot \mathbf{a}_2 + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} = -1 \\
\cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\
\cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\
\end{cases} \\
\Rightarrow && 2 (\mathbf{a}_1 \cdot \mathbf{a}_2) - 2(\mathbf{a}_3 \cdot \mathbf{a}_4) = 0
\end{align*}
Rather than adding the first two and last two, we could have done any pair, resulting in the relations:
\begin{align*}
\mathbf{a}_1 \cdot \mathbf{a}_2 &= \mathbf{a}_3 \cdot \mathbf{a}_4 \\
\mathbf{a}_1 \cdot \mathbf{a}_3 &= \mathbf{a}_2 \cdot \mathbf{a}_4 \\
\mathbf{a}_1 \cdot \mathbf{a}_4 &= \mathbf{a}_2 \cdot \mathbf{a}_3
\end{align*}
\begin{enumerate}
\item[(a)] The shape must be a parallelogram (from the angle requirement, but also cyclic quadrilateral (since all vectors are unit length), therefore it must be a rectangle
\item[(b)] Given it's a regular tetrahedron, $\mathbf{a}_i \cdot \mathbf{a}_j$ must be the same for all $i \neq j$, ie $-\frac13$.
We are interested in $|\mathbf{a}_i - \mathbf{a}_j|$ so consider,
\begin{align*}
|\mathbf{a}_i - \mathbf{a}_j|^2 &= (\mathbf{a}_i - \mathbf{a}_j) \cdot (\mathbf{a}_i - \mathbf{a}_j) \\
&= \mathbf{a}_i \cdot \mathbf{a}_i - 2 \mathbf{a}_i \cdot \mathbf{a}_j + \mathbf{a}_j \cdot \mathbf{a}_j \\
&= 1 - \frac23 + 1 \\
&= \frac43
\end{align*}
Therefore the unit side lengths are $\frac{2}{\sqrt{3}}$
\end{enumerate}
\end{questionparts}
This was the least popular of the Pure questions. Good solutions to this question often included clear diagrams to enable the angles being discussed to be identified easily. Many of the candidates were able to calculate the value of a·b correctly, but often did not fully justify that the triangle ABC was equilateral. For the second part, many candidates were again able to establish the relationship between scalar products, but less success was seen in identifying the type of quadrilateral. In the final part there were a large number of different approaches taken and many of these were completed successfully by some of the candidates.