A2 June 2022 Paper 2 Q1
1
(a) Find a vector which is perpendicular to both \(3\mathbf{i} - 5\mathbf{j} - \mathbf{k}\) and \(\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\). [1]
The equations of two lines are \(\mathbf{r} = 2\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} + \lambda(\mathbf{i} - 2\mathbf{j} + \mathbf{k})\) and \(\mathbf{r} = \mathbf{i} + 11\mathbf{j} - 4\mathbf{k} + \mu(-\mathbf{i} + 3\mathbf{j} - 2\mathbf{k})\).
(b) Show that the lines intersect, stating the point of intersection. [5]
| Scheme | Marks | AO |
|---|---|---|
| \((3\mathbf{i} - 5\mathbf{j} - \mathbf{k}) \times (\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}) = 23\mathbf{i} + 11\mathbf{j} + 14\mathbf{k}\) | B1 | 1.1 |
| [1] |
Notes
B1: or any non-zero multiple. ISW
| Scheme | Marks | AO |
|---|---|---|
| \(x\): \(2 + \lambda = 1 - \mu\) \(y\): \(3 - 2\lambda = 11 + 3\mu\) or \(z\): \(3 + \lambda = -4 - 2\mu\) | M1 | 1.1 |
| eg \(4 + 2\lambda = 2 - 2\mu \Rightarrow 7 = 13 + \mu\) | M1 | 1.1 |
| \(\mu = -6,\ \lambda = 5\) | A1 | 1.1 |
| eg \(z\): LHS \(= 3 + 5 = 8\) RHS \(= -4 - 2 \times (-6) = -4 - (-12) = 8\) | B1FT | 1.1 |
| PoI is \((7, -7, 8)\) | B1 | 1.1 |
| [5] |
Notes
M1: Any correct numeric equation
M1: Using any 2 equations to eliminate either \(\lambda\) or \(\mu\).
Must indicate which equations are used to find \(\lambda\) and \(\mu\).
B1FT: Check in unused (or all) equation(s)
FT their values of \(\lambda\) and \(\mu\) provided that LHS = RHS.
Calculations must be correct and answer must be evaluated.
LHS = 8, RHS = 8 is insufficient; some substitution must be evident.
B1: Condone as (position) vector