Higher November 2021 Paper 3 Q27
27 In the diagram
\(\overrightarrow{DE}\) \(= \mathbf{a}\)
\(\overrightarrow{DH}\) \(= \mathbf{b}\)
\(\overrightarrow{HG}\) \(= 8\mathbf{b}\)
\(EX : XH = 3 : 1\)
\(EF : FG = 1 : 3\)
Not drawn accurately
(a) Show that \(\quad\) \(\overrightarrow{DX}\) \(= \dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\) [2 marks]
(b) Is \(DXF\) a straight line?
Show working to support your answer. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: \(DH + HX\) | ||
| \(HE = \mathbf{a} - \mathbf{b}\) | M1 | implied by \(HX = \dfrac{1}{4}\mathbf{a} - \dfrac{1}{4}\mathbf{b}\) |
| \(\left(\mathbf{b} + \dfrac{1}{4}(\mathbf{a} - \mathbf{b}) =\right)\ \mathbf{b} + \dfrac{1}{4}\mathbf{a} - \dfrac{1}{4}\mathbf{b}\) \(= \dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\) | A1 | |
| Alternative method 2: \(DE + EX\) | ||
| \(EH = \mathbf{b} - \mathbf{a}\) | M1 | implied by \(EX = \dfrac{3}{4}\mathbf{b} - \dfrac{3}{4}\mathbf{a}\) |
| \(\left(\mathbf{a} + \dfrac{3}{4}(\mathbf{b} - \mathbf{a}) =\right)\ \mathbf{a} + \dfrac{3}{4}\mathbf{b} - \dfrac{3}{4}\mathbf{a}\) \(= \dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\) | A1 | |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: \(DF\) from \(DE + EF = DE + \dfrac{1}{4}EG\) | ||
| \((EG =)\ -\mathbf{a} + 9\mathbf{b}\) or \((EF =)\ -\dfrac{1}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) | M1 | oe |
| \((EF =)\ -\dfrac{1}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) and \((DF =)\ \mathbf{a} - \dfrac{1}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) | M1 | oe |
| \((DF =)\ \dfrac{3}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) | A1 | |
| \((DF =)\ 3\left(\dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right)\) and Yes | A1 | oe using a different correct scalar multiple for \(DF\) and \(DX\) |
| Alternative method 2: \(DF\) from \(DG + GF = DG + \dfrac{3}{4}GE\) | ||
| \((GE =)\ -9\mathbf{b} + \mathbf{a}\) or \((GF =)\ -\dfrac{27}{4}\mathbf{b} + \dfrac{3}{4}\mathbf{a}\) | M1 | oe |
| \((GF =)\ -\dfrac{27}{4}\mathbf{b} + \dfrac{3}{4}\mathbf{a}\) and \((DF =)\ 9\mathbf{b} - \dfrac{27}{4}\mathbf{b} + \dfrac{3}{4}\mathbf{a}\) | M1 | oe |
| \((DF =)\ \dfrac{3}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) | A1 | |
| \((DF =)\ 3\left(\dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right)\) and Yes | A1 | oe using a different correct scalar multiple for \(DF\) and \(DX\) |
| Alternative method 3: \(XF\) from \(XE + EF = \dfrac{3}{4}HE + \dfrac{1}{4}EG\) | ||
| \((XE =)\ \dfrac{3}{4}\mathbf{a} - \dfrac{3}{4}\mathbf{b}\) or \((EF =)\ -\dfrac{1}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) | M1 | oe |
| \((XF =)\ \dfrac{3}{4}\mathbf{a} - \dfrac{3}{4}\mathbf{b} - \dfrac{1}{4}\mathbf{a} + \dfrac{9}{4}\mathbf{b}\) | M1 | oe |
| \((XF =)\ \dfrac{2}{4}\mathbf{a} + \dfrac{6}{4}\mathbf{b}\) or \((XF =)\ \dfrac{1}{2}\mathbf{a} + \dfrac{3}{2}\mathbf{b}\) | A1 | |
| \((XF =)\ 2\left(\dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right)\) and Yes | A1 | oe using a different correct scalar multiple for \(XF\) and \(DX\) |
| Alternative method 4: \(XF\) from \(XH + HG + GF = \dfrac{1}{4}EH + HG + \dfrac{3}{4}GE\) | ||
| \((XH =)\ -\dfrac{1}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\) or \((GF =)\ -\dfrac{27}{4}\mathbf{b} + \dfrac{3}{4}\mathbf{a}\) | M1 | oe |
| \((XF =)\ -\dfrac{1}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b} + 8\mathbf{b} - \dfrac{27}{4}\mathbf{b} + \dfrac{3}{4}\mathbf{a}\) | M1 | oe |
| \((XF =)\ \dfrac{2}{4}\mathbf{a} + \dfrac{6}{4}\mathbf{b}\) or \((XF =)\ \dfrac{1}{2}\mathbf{a} + \dfrac{3}{2}\mathbf{b}\) | A1 | |
| \((XF =)\ 2\left(\dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right)\) and Yes | A1 | oe using a different correct scalar multiple for \(XF\) and \(DX\) |
Additional guidance
Method marks may be awarded for correct work seen on diagram or in working, with no or incorrect answer, even if this is seen amongst multiple attempts