A2 June 2023 Paper 1 Q9
9 The position vectors of the points \(A\), \(B\) and \(C\) are
\[\begin{aligned} \mathbf{a} &= 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \\ \mathbf{b} &= -\mathbf{i} - 8\mathbf{j} + 2\mathbf{k} \\ \mathbf{c} &= -2\mathbf{j} \end{aligned}\]respectively.
(a) Find the area of the triangle \(ABC\) [4 marks]
(b) The points \(A\), \(B\) and \(C\) all lie in the plane \(\Pi\)
Find an equation of the plane \(\Pi\), in the form \(\mathbf{r} \bullet \mathbf{n} = d\) [2 marks]
(c) The point \(P\) has position vector \(\mathbf{p} = \mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\)
Find the exact distance of \(P\) from \(\Pi\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains two vectors in the plane of the triangle | B1 | 1.1b |
| Selects a method to find the area of a triangle for example by taking the vector product of their two vectors | M1 | 3.1a |
| Uses a correct formula for the area of a triangle | M1 | 1.2 |
| Obtains the correct area with no incorrect working | A1 | 1.1b |
| (4) |
Typical solution
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{bmatrix} -3 \\ -9 \\ 0 \end{bmatrix} \times \begin{bmatrix} -2 \\ -3 \\ -2 \end{bmatrix} = \begin{bmatrix} 18 \\ -6 \\ -9 \end{bmatrix}\]\[\begin{aligned} \text{Area} &= \frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \frac{1}{2}\sqrt{18^2 + (-6)^2 + (-9)^2} \\ &= \frac{21}{2} \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Forms the scalar product of their normal and a position vector of a point in \(\Pi\) | M1 | 1.1a |
| Obtains a correct equation of \(\Pi\) | A1 | 1.1b |
| (2) |
Typical solution
\[d = \begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix} \bullet \begin{bmatrix} 6 \\ -2 \\ -3 \end{bmatrix} = 4\]\[\mathbf{r} \bullet \begin{bmatrix} 6 \\ -2 \\ -3 \end{bmatrix} = 4\]| Scheme | Marks | AO |
|---|---|---|
| Selects a method to find the parallel plane that \(P\) lies in by using the scalar product of their normal vector and \(P\) to obtain their “\(-8\)” or Uses a correct formula to find the required distance or Substitutes equation of perpendicular line from \(P\) to \(\Pi\) into equation of \(\Pi\) | M1 | 3.1a |
| Uses the magnitude of their normal vector to divide their “4” and “\(-8\)” in order to find the distance of \(P\) from \(\Pi\) or Substitutes correctly into a formula for the required distance | M1 | 2.2a |
| Obtains the correct distance of \(P\) from \(\Pi\) | A1 | 1.1b |
| (3) | ||
| (9 marks) |