FP3 June 2014 Q8
8. The position vectors of the points \(A\), \(B\) and \(C\) from a fixed origin \(O\) are \[\mathbf{a} = \mathbf{i} - \mathbf{j}, \quad \mathbf{b} = \mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \mathbf{c} = 2\mathbf{j} + \mathbf{k}\] respectively.
(a) Using vector products, find the area of the triangle \(ABC\). (4)
(b) Show that \(\dfrac{1}{6}\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0\) (3)
(c) Hence or otherwise, state what can be deduced about the vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\). (1)
| Scheme | Marks |
|---|---|
| \(\mathbf{OA} = \begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}, \mathbf{OB} = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, \mathbf{OC} = \mathbf{AB} = \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}, \mathbf{BC} = \begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}, \mathbf{AC} = \begin{pmatrix}-1 \\ 3 \\ 1\end{pmatrix}\) | |
| \(\mathbf{AB} \times \mathbf{AC} = -\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) Or e.g. \(\mathbf{BA} \times \mathbf{BC} = \mathbf{i} + \mathbf{j} - 2\mathbf{k}\) M1: Attempt vector product for two sides of the triangle. If the method is unclear, at least 2 components must be correct. A1: Correct vector | M1A1 |
| Area ABC \(= \dfrac{1}{2}\sqrt{1^2 + 1^2 + 2^2}\) Attempts \(\dfrac{1}{2}|\text{their}\,\mathbf{AB} \times \mathbf{AC}|\) Dependent on the first M | dM1 |
| \(\dfrac{1}{2}\sqrt{6}\) Accept equivalents or awrt 1.22 | A1 |
| Note that triangles OAB and OBC have the same area but score 0/4 It must be triangle ABC | |
| (4) |
| Scheme | Marks |
|---|---|
| \(\mathbf{b} \times \mathbf{c} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{j} + \mathbf{k}) = -\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) Attempt \(\mathbf{b} \times \mathbf{c}\). If the method is unclear, at least 2 components must be correct. | M1 |
| \(= \left(\dfrac{1}{6}\right)(\mathbf{i} - \mathbf{j}) \cdot (-\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = \left(\dfrac{1}{6}\right)(-1 + 1) = 0\) M1: Attempt scalar product of \(\mathbf{a}\) with their \(\mathbf{b} \times \mathbf{c}\) to obtain a number not a vector. A1: Obtains = 0 with no errors (allow omission of \(\tfrac{1}{6}\) for all 3 marks) Just \(= \mathbf{a} \cdot (-\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 0\) would lose the A1 | M1A1 |
| (3) |
Notes
Alternative
| Scheme | Marks |
|---|---|
| \((\mathbf{a} \cdot \mathbf{b} \times \mathbf{c} =)\begin{vmatrix}1 & -1 & 0 \\ 1 & 1 & 1 \\ 0 & 2 & 1\end{vmatrix}\) Writes this statement (allow other brackets provided the determinant is implied later) | M1 |
| \(= (1 - 2) + 1(1) - 0 = 0\) M1: Clear attempt at determinant A1: Obtains = 0 with no errors (allow omission of \(\tfrac{1}{6}\) for all 3 marks) | M1A1 |
| Scheme | Marks |
|---|---|
| Volume of tetrahedron (OABC) = 0 \(\mathbf{a} = \mathbf{b} - \mathbf{c}\) oe or \(\mathbf{c} = \mathbf{b} - \mathbf{a}\) oe \(\mathbf{b} \times \mathbf{c}\) is perpendicular to \(\mathbf{a}\) or \(\mathbf{a}\) is parallel to \(\mathbf{CB}\) All vectors/points lie in the same plane OABC is a parallelogram \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) are linearly dependent Do not isw – if there are contradictory or wrong statements award B0 | B1 |
| (1) | |
| (8 marks) |