Edexcel Paper 1 Specimen — Question 7 5 marks

Exam BoardEdexcel
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks5
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Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeAngle between two vectors/lines (direct)
DifficultyStandard +0.3 This is a straightforward application of the scalar product formula to find an angle in a triangle. Students need to find AC using vector addition, then apply the standard formula cos(θ) = (a·b)/(|a||b|). While it requires multiple steps (vector addition, two magnitude calculations, dot product, inverse cosine), these are all routine procedures with no conceptual difficulty or problem-solving insight required. Slightly easier than average due to its mechanical nature.
Spec1.07n Stationary points: find maxima, minima using derivatives1.09a Sign change methods: locate roots1.09d Newton-Raphson method
7. Figure 2 Figure 2 shows a sketch of a triangle \(A B C\).
Given \(\overrightarrow { A B } = 2 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }\) and \(\overrightarrow { B C } = \mathbf { i } - 9 \mathbf { j } + 3 \mathbf { k }\),
show that \(\angle B A C = 105.9 ^ { \circ }\) to one decimal place.
Question 7:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} = 2\mathbf{i}+3\mathbf{j}+\mathbf{k}+\mathbf{i}-9\mathbf{j}+3\mathbf{k} = 3\mathbf{i}-6\mathbf{j}+4\mathbf{k}\)M1 Attempts to find \(\overrightarrow{AC}\) using \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\)
Attempts to find any one length using 3-D PythagorasM1
\(AB =\sqrt{14},\quad
\(\cos BAC = \frac{14+61-91}{2\sqrt{14}\sqrt{61}}\)M1 Uses cosine rule; also allow \(\cos BAC = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{
Angle \(BAC = 105.9°\)A1* Show that — all aspects must be correct 
## Question 7:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} = 2\mathbf{i}+3\mathbf{j}+\mathbf{k}+\mathbf{i}-9\mathbf{j}+3\mathbf{k} = 3\mathbf{i}-6\mathbf{j}+4\mathbf{k}$ | M1 | Attempts to find $\overrightarrow{AC}$ using $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$ |
| Attempts to find any one length using 3-D Pythagoras | M1 | |
| $|AB|=\sqrt{14},\quad |AC|=\sqrt{61},\quad |BC|=\sqrt{91}$ | A1ft | Finds all three lengths; follow through on their $|AC|$ |
| $\cos BAC = \frac{14+61-91}{2\sqrt{14}\sqrt{61}}$ | M1 | Uses cosine rule; also allow $\cos BAC = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|AB||AC|}$ |
| Angle $BAC = 105.9°$ | A1* | Show that — all aspects must be correct |
7.

Figure 2

Figure 2 shows a sketch of a triangle $A B C$.\\
Given $\overrightarrow { A B } = 2 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }$ and $\overrightarrow { B C } = \mathbf { i } - 9 \mathbf { j } + 3 \mathbf { k }$,\\
show that $\angle B A C = 105.9 ^ { \circ }$ to one decimal place.

\hfill \mbox{\textit{Edexcel Paper 1  Q7 [5]}}
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