June 2023 Paper 2 Q2
2 The points \(O\) and \(A\) have position vectors \(\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 6 \\ 0 \\ 8 \end{pmatrix}\) respectively. The point \(P\) is such that \(\overrightarrow{OP} = k\overrightarrow{OA}\), where \(k\) is a non-zero constant.
Point \(B\) has position vector \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and angle \(OPB\) is a right angle.
| Scheme | Marks | AO |
|---|---|---|
| \(10|k|\) | B1 | 1.1 |
| [1] |
Notes
B1: Allow \(10k\)
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{BP} = \begin{pmatrix} 6k - 1 \\ -2 \\ 8k - 3 \end{pmatrix}\) or \(\overrightarrow{PB} = \begin{pmatrix} 1 - 6k \\ 2 \\ 3 - 8k \end{pmatrix}\) | B1 | 3.1a |
| \(100k^2 + (6k - 1)^2 + (-2)^2 + (8k - 3)^2 = 14\) | M1 | 1.1 |
| \(200k^2 - 60k + 14 = 14\) | A1 | 1.1 |
| \(k = \dfrac{3}{10}\) | A1 | 1.1 |
| [4] |
Notes
B1: May be implied (but must be fully correct to imply, i.e. including (\(-2\)))
M1: Attempt \(OP^2 + BP^2 = OB^2\) FT their \(OP\) and \(\overrightarrow{BP}\) or \(\overrightarrow{PB}\)
A1: Correct equation after expanding brackets
A1: oe Condone inclusion of \(k = 0\) (whether eliminated or not)
Alternative method using scalar product
| Scheme | Marks |
|---|---|
| \(\overrightarrow{BP} = \begin{pmatrix} 6k - 1 \\ -2 \\ 8k - 3 \end{pmatrix}\) or \(\overrightarrow{PB} = \begin{pmatrix} 1 - 6k \\ 2 \\ 3 - 8k \end{pmatrix}\) | B1 |
| \(6k(6k - 1) + 0 + 8k(8k - 3) = 0\) | M1 |
| \(100k^2 - 30k = 0\) | A1 |
| \(k = \dfrac{3}{10}\) | A1 |
B1: May be implied
M1: Attempt \(\overrightarrow{OP} \cdot \overrightarrow{BP} = 0\) FT their \(\overrightarrow{OP}\) and \(\overrightarrow{BP}\) or \(\overrightarrow{PB}\)
(or attempt \(\overrightarrow{OA} \cdot \overrightarrow{BP} = 0 \Rightarrow 6(6k - 1) + 0 + 8(8k - 3) = 0\))
Must be algebraic – i.e. an equation of this form in \(k\).
A1: Correct equation after expanding brackets
Or \(100k - 30 = 0\)
A1: oe Condone inclusion of \(k = 0\) (whether eliminated or not)