| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of triangle from given side vectors or coordinates |
| Difficulty | Standard +0.3 This is a straightforward C4 vectors question testing standard techniques: dot product for angle, cross product (or sine formula) for area, and perpendicular distance formula. All three parts follow directly from learned formulas with routine calculation, requiring no problem-solving insight beyond recognizing which formula to apply. Slightly above average difficulty only due to the three-part structure and exact form requirement. |
| Spec | 1.10c Magnitude and direction: of vectors4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{ | 1 \times 6 + 5 \times 3 + (-1) \times (-6) | }{\sqrt{1 + 25 + 1} \times \sqrt{36 + 9 + 36}}\) |
| \(= \frac{27}{\sqrt{27} \times \sqrt{81}} = \frac{\sqrt{27}}{9} = \frac{3\sqrt{3}}{9} = \frac{1}{3}\sqrt{3}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\sin(\angle AOB) = \sqrt{1 - \left(\frac{1}{3}\sqrt{3}\right)^2} = \sqrt{\frac{2}{3}}\) | M1 A1 |
| \(\text{area} = \frac{1}{2} \times 3\sqrt{3} \times 9 \times \sqrt{\frac{2}{3}} = \frac{27\sqrt{2}}{2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= OA \times \sin(\angle AOB) = 3\sqrt{3} \times \sqrt{\frac{2}{3}} = 3\sqrt{2}\) | M1 A1 | (10) |
**(a)**
$= \frac{|1 \times 6 + 5 \times 3 + (-1) \times (-6)|}{\sqrt{1 + 25 + 1} \times \sqrt{36 + 9 + 36}}$ | M1 A1 |
$= \frac{27}{\sqrt{27} \times \sqrt{81}} = \frac{\sqrt{27}}{9} = \frac{3\sqrt{3}}{9} = \frac{1}{3}\sqrt{3}$ | M1 A1 |
**(b)**
$\sin(\angle AOB) = \sqrt{1 - \left(\frac{1}{3}\sqrt{3}\right)^2} = \sqrt{\frac{2}{3}}$ | M1 A1 |
$\text{area} = \frac{1}{2} \times 3\sqrt{3} \times 9 \times \sqrt{\frac{2}{3}} = \frac{27\sqrt{2}}{2}$ | M1 A1 |
**(c)**
$= OA \times \sin(\angle AOB) = 3\sqrt{3} \times \sqrt{\frac{2}{3}} = 3\sqrt{2}$ | M1 A1 | (10)Relative to a fixed origin, $O$, the points $A$ and $B$ have position vectors $\begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 3 \\ -6 \end{pmatrix}$ respectively.
Find, in exact, simplified form,
\begin{enumerate}[label=(\alph*)]
\item the cosine of $\angle AOB$, [4]
\item the area of triangle $OAB$, [4]
\item the shortest distance from $A$ to the line $OB$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [10]}}