OCR MEI C4 — Question 3 6 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeArea of triangle from given side vectors or coordinates
DifficultyStandard +0.3 This is a straightforward application of the scalar product to verify perpendicularity, followed by a simple area calculation using the right-angle triangle formula. Both techniques are standard C4 content requiring only routine execution of learned methods with no problem-solving insight needed.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles
3 A triangle ABC has vertices \(\mathrm { A } ( - 2,4,1 ) , \mathrm { B } ( 2,3,4 )\) and \(\mathrm { C } ( 4,8,3 )\). By calculating a suitable scalar product, show that angle ABC is a right angle. Hence calculate the area of the triangle. [6]
Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{BA} = \begin{pmatrix}-4\\1\\-3\end{pmatrix},\ \overrightarrow{BC} = \begin{pmatrix}2\\5\\-1\end{pmatrix}\)B1 soi, condone wrong sense
\(\overrightarrow{BA}\cdot\overrightarrow{BC} = \begin{pmatrix}-4\\1\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\5\\-1\end{pmatrix} = (-4)\times2 + 1\times5 + (-3)\times(-1)\)M1 scalar product
\(= -8 + 5 + 3 = 0 \Rightarrow\) angle \(ABC = 90°\)A1 \(= 0\)
Area of triangle \(= \frac{1}{2} \times BA \times BC\)M1 area of triangle formula oe
\(= \frac{1}{2}\times\sqrt{(-4)^2+1^2+3^2}\times\sqrt{2^2+5^2+(-1)^2}\)M1 length formula
\(= \frac{1}{2}\times\sqrt{26}\times\sqrt{30} = 13.96\) sq unitsA1 [6] accept \(14.0\) and \(\sqrt{195}\) 
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{BA} = \begin{pmatrix}-4\\1\\-3\end{pmatrix},\ \overrightarrow{BC} = \begin{pmatrix}2\\5\\-1\end{pmatrix}$ | B1 | soi, condone wrong sense |
| $\overrightarrow{BA}\cdot\overrightarrow{BC} = \begin{pmatrix}-4\\1\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\5\\-1\end{pmatrix} = (-4)\times2 + 1\times5 + (-3)\times(-1)$ | M1 | scalar product |
| $= -8 + 5 + 3 = 0 \Rightarrow$ angle $ABC = 90°$ | A1 | $= 0$ |
| Area of triangle $= \frac{1}{2} \times BA \times BC$ | M1 | area of triangle formula oe |
| $= \frac{1}{2}\times\sqrt{(-4)^2+1^2+3^2}\times\sqrt{2^2+5^2+(-1)^2}$ | M1 | length formula |
| $= \frac{1}{2}\times\sqrt{26}\times\sqrt{30} = 13.96$ sq units | A1 [6] | accept $14.0$ and $\sqrt{195}$ |
3 A triangle ABC has vertices $\mathrm { A } ( - 2,4,1 ) , \mathrm { B } ( 2,3,4 )$ and $\mathrm { C } ( 4,8,3 )$. By calculating a suitable scalar product, show that angle ABC is a right angle. Hence calculate the area of the triangle. [6]

\hfill \mbox{\textit{OCR MEI C4  Q3 [6]}}
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