FP3 June 2012 Q3
3. The position vectors of the points \(A\), \(B\) and \(C\) relative to an origin \(O\) are \(\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}\), \(7\mathbf{i} - 3\mathbf{k}\) and \(4\mathbf{i} + 4\mathbf{j}\) respectively.
Find
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AC} = 3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\), \(\overrightarrow{BC} = -3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\) | B1, B1 |
| \(\overrightarrow{AC} \times \overrightarrow{BC} = 10\mathbf{i} - 15\mathbf{j} + 30\mathbf{k}\) | M1 A1 |
| (4) |
Notes
a1B1: \(\overrightarrow{AC} = 3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\) cao, any form
a2B1: \(\overrightarrow{BC} = -3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\) cao, any form
a1M1: Attempt to find cross product, modulus of one term correct.
a1A1: cao, any form.
| Scheme | Marks |
|---|---|
| Area of triangle \(ABC = \tfrac{1}{2}\left|10\mathbf{i} - 15\mathbf{j} + 30\mathbf{k}\right| = \tfrac{1}{2}\sqrt{1225} = 17.5\) | M1 A1 |
| (2) |
Notes
b1M1: modulus of their answer to (a) – condone missing ½ here. To finding area of triangle by correct method.
b1A1: cao.
| Scheme | Marks |
|---|---|
| Equation of plane is \(10x - 15y + 30z = -20\) or \(2x - 3y + 6z = -4\) | M1 |
| So \(\mathbf{r} \cdot (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = -4\) or correct multiple | A1 |
| (2) | |
| (8 marks) |
Notes
c1M1: [Using their answer to (a) to] find equation of plane. Look for \(\mathbf{a} \cdot \mathbf{n}\) or \(\mathbf{b} \cdot \mathbf{n}\) or \(\mathbf{c} \cdot \mathbf{n}\) for p.
c1A1: cao