| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle between two vectors/lines (direct) |
| Difficulty | Moderate -0.3 This is a straightforward vector question requiring basic vector addition and the scalar product formula for angles. Part (a) is simple vector arithmetic (AB + BC = AC), and part (b) applies the standard dot product formula with given vectors. The calculations are routine with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = -3\mathbf{i}-4\mathbf{j}-5\mathbf{k}+\mathbf{i}+\mathbf{j}+4\mathbf{k}\) | M1 | Must attempt to add not subtract. If no method shown, implied by two correct components |
| \(= -2\mathbf{i}-3\mathbf{j}-\mathbf{k}\) | A1 | Correct vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((AC^2) = 2^2+3^2+1^2\), \((AB^2) = 3^2+4^2+5^2\), \((BC^2) = 1^2+1^2+4^2\) | M1 | At least 2 of these correct. Follow through on their \(\overrightarrow{AC}\) |
| \(2^2+3^2+1^2 = 3^2+4^2+5^2+1^2+1^2+4^2-2\sqrt{3^2+4^2+5^2}\sqrt{1^2+1^2+4^2}\cos ABC\) | M1 | Correct application of cosine rule. Condone slips on lengths but sides must be in correct position |
| \(14 = 50+18-2\sqrt{50}\sqrt{18}\cos ABC \Rightarrow \cos ABC = \frac{50+18-14}{2\sqrt{50}\sqrt{18}} = \frac{9}{10}\) | A1* | Correct completion with sufficient intermediate working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2 = 3^2+4^2+5^2\), \(BC^2 = 1^2+1^2+4^2\) | M1 | Correct application of Pythagoras for sides \(AB\) and \(BC\) |
| \(\overrightarrow{BA}\cdot\overrightarrow{BC} = (3\mathbf{i}+4\mathbf{j}+5\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}+4\mathbf{k}) = 27 = \sqrt{3^2+4^2+5^2}\sqrt{1^2+1^2+4^2}\cos ABC\) | M1 | Recognises requirement for scalar product |
| \(27 = \sqrt{50}\sqrt{18}\cos ABC \Rightarrow \cos ABC = \frac{27}{\sqrt{50}\sqrt{18}} = \frac{9}{10}\) | A1* | Correct completion with sufficient intermediate working |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = -3\mathbf{i}-4\mathbf{j}-5\mathbf{k}+\mathbf{i}+\mathbf{j}+4\mathbf{k}$ | M1 | Must attempt to add not subtract. If no method shown, implied by two correct components |
| $= -2\mathbf{i}-3\mathbf{j}-\mathbf{k}$ | A1 | Correct vector |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(AC^2) = 2^2+3^2+1^2$, $(AB^2) = 3^2+4^2+5^2$, $(BC^2) = 1^2+1^2+4^2$ | M1 | At least 2 of these correct. Follow through on their $\overrightarrow{AC}$ |
| $2^2+3^2+1^2 = 3^2+4^2+5^2+1^2+1^2+4^2-2\sqrt{3^2+4^2+5^2}\sqrt{1^2+1^2+4^2}\cos ABC$ | M1 | Correct application of cosine rule. Condone slips on lengths but sides must be in correct position |
| $14 = 50+18-2\sqrt{50}\sqrt{18}\cos ABC \Rightarrow \cos ABC = \frac{50+18-14}{2\sqrt{50}\sqrt{18}} = \frac{9}{10}$ | A1* | Correct completion with sufficient intermediate working |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2 = 3^2+4^2+5^2$, $BC^2 = 1^2+1^2+4^2$ | M1 | Correct application of Pythagoras for sides $AB$ and $BC$ |
| $\overrightarrow{BA}\cdot\overrightarrow{BC} = (3\mathbf{i}+4\mathbf{j}+5\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}+4\mathbf{k}) = 27 = \sqrt{3^2+4^2+5^2}\sqrt{1^2+1^2+4^2}\cos ABC$ | M1 | Recognises requirement for scalar product |
| $27 = \sqrt{50}\sqrt{18}\cos ABC \Rightarrow \cos ABC = \frac{27}{\sqrt{50}\sqrt{18}} = \frac{9}{10}$ | A1* | Correct completion with sufficient intermediate working |
---6.
Figure 1
Figure 1 shows a sketch of triangle $A B C$.\\
Given that
\begin{itemize}
\item $\overrightarrow { A B } = - 3 \mathbf { i } - 4 \mathbf { j } - 5 \mathbf { k }$
\item $\overrightarrow { B C } = \mathbf { i } + \mathbf { j } + 4 \mathbf { k }$
\begin{enumerate}[label=(\alph*)]
\item find $\overrightarrow { A C }$
\item show that $\cos A B C = \frac { 9 } { 10 }$
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q6 [5]}}