SPS SPS FM Pure 2023 June — Question 5 5 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2023
SessionJune
Marks5
TopicVectors: Cross Product & Distances
TypeAngle between vectors using scalar product
DifficultyStandard +0.3 This is a straightforward vector geometry problem requiring calculation of angle using the scalar product formula. Students need to find AC, compute two dot products, apply cos θ = (a·b)/(|a||b|), and use inverse cosine. While it involves multiple steps (finding vectors, magnitudes, dot product, angle), these are all standard Further Maths techniques with no conceptual difficulty or novel insight required. Slightly easier than average due to its routine algorithmic nature.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication

\includegraphics{figure_5} Figure 1 shows a sketch of a triangle \(ABC\). Given \(\overrightarrow{AB} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\) and \(\overrightarrow{BC} = \mathbf{i} - 9\mathbf{j} + 3\mathbf{k}\), show that \(\angle BAC = 105.9°\) to one decimal place. [5]

\includegraphics{figure_5}

Figure 1 shows a sketch of a triangle $ABC$.

Given $\overrightarrow{AB} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ and $\overrightarrow{BC} = \mathbf{i} - 9\mathbf{j} + 3\mathbf{k}$,

show that $\angle BAC = 105.9°$ to one decimal place. [5]

\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q5 [5]}}