A2 June 2019 Paper 2 Q7
7 The points \(A\), \(B\) and \(C\) have coordinates \(A(4, 5, 2)\), \(B(-3, 2, -4)\) and \(C(2, 6, 1)\)
(a) Use a vector product to show that the area of triangle \(ABC\) is \(\dfrac{5\sqrt{11}}{2}\) [4 marks]
(b) The points \(A\), \(B\) and \(C\) lie in a plane.
Find a vector equation of the plane in the form \(\mathbf{r}.\mathbf{n} = k\) [1 mark]
(c) Hence find the exact distance of the plane from the origin. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Forms two vectors from A, B and C, at least one correct | M1 | 1.1a |
| Obtains the correct vector product. | A1 | 1.1b |
| Uses their vector product correctly in a formula for the area of a triangle | M1 | 1.2 |
| Uses a rigorous argument to show the required result, including stating “Area =” oe | R1 | 2.1 |
Typical solution
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} -7 \\ -3 \\ -6 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ -13 \end{pmatrix}\]\[\text{Area} = \frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|\]\[= \frac{\sqrt{9^2 + 5^2 + (-13)^2}}{2} = \frac{5\sqrt{11}}{2}\]| Scheme | Marks | AO |
|---|---|---|
| States the correct equation | B1 | 1.1b |
Typical solution
\[\mathbf{r}.\begin{pmatrix} 9 \\ 5 \\ -13 \end{pmatrix} = 35\]| Scheme | Marks | AO |
|---|---|---|
| Divides their non-zero \(k\) by the magnitude of their normal vector to obtain their exact value | B1F | 3.1a |
| (6 marks) |