Year: 2024
Paper: 2
Question Number: 4
Course: UFM Pure
Section: Vectors
No solution available for this problem.
Many candidates produced good solutions to the questions, with the majority of candidates opting to focus on the pure questions of the paper. Candidates demonstrated very good ability, particularly in the area of manipulating algebra. Many candidates produced clear diagrams which in many cases meant that they were more successful in their attempts at their questions than those who did not do so. The paper also contained a number of places where the answer to be reached was given in the question. In such cases, candidates must be careful to ensure that they provide sufficient evidence of the method used to reach the result in order to gain full credit.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, if $O$, $C$ and $D$ are non-collinear points in three dimensional space, we will call the non-zero vector $\mathbf{v}$ a \emph{bisecting vector} for angle $COD$ if $\mathbf{v}$ lies in the plane $COD$, the angle between $\mathbf{v}$ and $\overrightarrow{OC}$ is equal to the angle between $\mathbf{v}$ and $\overrightarrow{OD}$, and both angles are less than $90^\circ$.
\begin{questionparts}
\item Let $O$, $X$ and $Y$ be non-collinear points in three-dimensional space, and define $\mathbf{x} = \overrightarrow{OX}$ and $\mathbf{y} = \overrightarrow{OY}$.
Let $\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}$.
\begin{enumerate}
\item Show that $\mathbf{b}$ is a bisecting vector for angle $XOY$.
Explain, using a diagram, why any other bisecting vector for angle $XOY$ is a positive multiple of $\mathbf{b}$.
\item Find the value of $\lambda$ such that the point $B$, defined by $\overrightarrow{OB} = \lambda\mathbf{b}$, lies on the line $XY$. Find also the ratio in which the point $B$ divides $XY$.
\item Show, in the case when $OB$ is perpendicular to $XY$, that the triangle $XOY$ is isosceles.
\end{enumerate}
\item Let $O$, $P$, $Q$ and $R$ be points in three-dimensional space, no three of which are collinear. A bisecting vector is chosen for each of the angles $POQ$, $QOR$ and $ROP$. Show that the three angles between them are either all acute, all obtuse or all right angles.
\end{questionparts}
About one third of candidates attempted this question. Many of the attempts struggled to explain the reasoning sufficiently clearly, often missing one or two important details. In part (i) most candidates were able to prove that the cosines of the two angles were equal, but did not justify details such as checking that everything lies in the same plane. The most problematic part of this part of the question for candidates was (i)(b). The most successful approach was to use the fact that XB and BY lie on the same line and then use algebra to calculate the value of λ and the ratio. A small number of candidates were able to achieve full marks with geometric arguments, but most such attempts did not give sufficient justification to be fully convincing. Candidates often produced successful attempts at (i)(c), although again some failed to justify elements of the solution fully enough in some parts. Part (ii) was found to be harder than the rest of the question. The majority who attempted this part correctly identified the dot product to be considered, but most did not recognize the symmetry within the three expressions.