June 2025 Paper 2 Q17
17 A triangle has vertices \(A\), \(B\) and \(C\) with position vectors given by
\[\overrightarrow{OA} = \begin{bmatrix}-3\\5\\1\end{bmatrix},\ \overrightarrow{OB} = \begin{bmatrix}5\\4\\-2\end{bmatrix} \text{ and } \overrightarrow{OC} = \begin{bmatrix}1\\-1\\0\end{bmatrix}\](a)
(i) Find \(\overrightarrow{AB}\) [1 mark]
(ii) Find the magnitude of \(\overrightarrow{AB}\) [1 mark]
(b) Hence show that triangle \(ABC\) is a scalene triangle. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(\begin{bmatrix}8\\-1\\-3\end{bmatrix}\) ACF | B1 | 1.1b |
| (1) | ||
| (ii) Finds the magnitude of their \(\overrightarrow{AB}\) AWRT 8.6 | B1F | 1.1b |
| (1) |
Typical solution
(a)(i)
\[\overrightarrow{AB} = \begin{bmatrix}8\\-1\\-3\end{bmatrix}\](a)(ii)
\[\left|AB\right| = \sqrt{64 + 1 + 9} = \sqrt{74}\]| Scheme | Marks | AO |
|---|---|---|
| Uses vector geometry to work out \(\overrightarrow{OC} - \overrightarrow{OA}\) or \(\overrightarrow{OC} - \overrightarrow{OB}\) OE PI \(\pm\begin{bmatrix}4\\-6\\-1\end{bmatrix}\) or \(\pm\begin{bmatrix}-4\\-5\\2\end{bmatrix}\) Condone missing arrows Condone missing or incorrect labels | M1 | 1.1a |
| Obtain the correct vectors for \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\) and \(\overrightarrow{BC}\) or \(\overrightarrow{CB}\) The vectors must be correctly labelled Condone missing arrows | A1 | 1.1b |
| Find the magnitude of their \(\overrightarrow{AC}\) or \(\overrightarrow{BC}\) OE PI \(\sqrt{53}\) or \(3\sqrt{5}\) OE AWRT 7.3 or 6.7 If \(3\sqrt{5}\) is obtained from \(\overrightarrow{OB}\) score M0 | M1 | 3.1a |
| Obtains \(\sqrt{53}\) and \(3\sqrt{5}\) OE from correct vectors. AWRT 7.3 and 6.7 | A1 | 1.1b |
| Completes reasoned argument to obtain the three lengths: \(\left|BC\right| = \sqrt{45}\) \(\left|AC\right| = \sqrt{53}\) OE \(\left|AB\right| = \sqrt{74}\) and Concludes that \(ABC\) is a scalene triangle with no incorrect reasoning. Must have scored M1A1M1A1 Must not equate a vector to a length. AG | R1 | 2.1 |
| (5) | ||
| (7 marks) |
Typical solution
\[\overrightarrow{AC} = \begin{bmatrix}1\\-1\\0\end{bmatrix} - \begin{bmatrix}-3\\5\\1\end{bmatrix} = \begin{bmatrix}4\\-6\\-1\end{bmatrix}\]\[\overrightarrow{BC} = \begin{bmatrix}1\\-1\\0\end{bmatrix} - \begin{bmatrix}5\\4\\-2\end{bmatrix} = \begin{bmatrix}-4\\-5\\2\end{bmatrix}\]\[\left|AB\right| = \sqrt{74}\]\[\left|BC\right| = \sqrt{16 + 25 + 4} = \sqrt{45}\]\[\left|AC\right| = \sqrt{53}\]All sides are different in length. Therefore, triangle \(ABC\) is a scalene triangle.