Higher June 2022 Paper 2R Q25
25 \(ABCD\) is a parallelogram and \(ADM\) is a straight line.
Diagram NOT accurately drawn
\(\overrightarrow{AB} = \mathbf{a}\) \(\overrightarrow{BC} = \mathbf{b}\) \(\overrightarrow{DM} = \dfrac{1}{2}\mathbf{b}\)
\(K\) is the point on \(AB\) such that \(AK : AB = \lambda : 1\)
\(L\) is the point on \(CD\) such that \(CL : CD = \mu : 1\)
\(KLM\) is a straight line.
Given that \(\lambda : \mu = 1 : 2\)
use a vector method to find the value of \(\lambda\) and the value of \(\mu\)
(5)
| Scheme | Marks |
|---|---|
| eg \(\overrightarrow{AK} = \lambda\mathbf{a}\) \(\overrightarrow{KB} = (1 - \lambda)\mathbf{a}\) \(\overrightarrow{CL} = -\mu\mathbf{a}\) \(\overrightarrow{DL} = (1 - \mu)\mathbf{a}\) or eg \(\overrightarrow{AK} = \tfrac{1}{2}\mu\mathbf{a}\) \(\overrightarrow{KB} = (1 - \tfrac{1}{2}\mu)\mathbf{a}\) \(\overrightarrow{CL} = -2\lambda\mathbf{a}\) \(\overrightarrow{DL} = (1 - 2\lambda)\mathbf{a}\) | M1 |
| eg \(\overrightarrow{KL} = -\lambda\mathbf{a} + \mathbf{b} + (1 - \mu)\mathbf{a}\) or \(= (1 - \lambda - \mu)\mathbf{a} + \mathbf{b}\) \(\overrightarrow{LM} = (\mu - 1)\mathbf{a} + 0.5\mathbf{b}\) \(\overrightarrow{KM} = -\lambda\mathbf{a} + \mathbf{b} + 0.5\mathbf{b}\;(= -\lambda\mathbf{a} + 1.5\mathbf{b})\) or eg \(\overrightarrow{KL} = \mathbf{b} + (1 - \tfrac{3}{2}\mu)\mathbf{a}\) or \(\overrightarrow{KL} = \mathbf{b} + (1 - 3\lambda)\mathbf{a}\) \(\overrightarrow{LM} = (2\lambda - 1)\mathbf{a} + \tfrac{1}{2}\mathbf{b}\) or \(\overrightarrow{KM} = -\lambda\mathbf{a} + \tfrac{3}{2}\mathbf{b}\) or | M1 |
| Two of the above – may have used \(2\lambda = \mu\) to write all in one of \(\lambda\) or \(\mu\) May be simplified or not – so may have brackets or not | M1 |
eg using \(\overrightarrow{KM} = -\lambda\mathbf{a} + 1.5\mathbf{b}\) and \(\overrightarrow{LM} = (2\lambda - 1)\mathbf{a} + \tfrac{1}{2}\mathbf{b}\) \(\overrightarrow{LM} = x\overrightarrow{KM}\) gives \(\dfrac{-\lambda x}{2\lambda - 1} = \dfrac{1.5x}{0.5} \Rightarrow 3.5\lambda = 1.5 \Rightarrow \lambda = \dfrac{3}{7}\) oe Answer: \(\lambda = \dfrac{3}{7}\) or \(\mu = \dfrac{6}{7}\) | A1 |
Working required Answer: \(\lambda = \dfrac{3}{7}\) & \(\mu = \dfrac{6}{7}\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for correctly using the ratio to form an expression for a vector passing through \(K\) or \(L\) could be in terms of \(\lambda\) or \(\mu\)
\(\overrightarrow{AK}\) or \(\overrightarrow{KA}\), \(\overrightarrow{KB}\) or \(\overrightarrow{BK}\), \(\overrightarrow{CL}\) or \(\overrightarrow{LC}\), \(\overrightarrow{DL}\) or \(\overrightarrow{LD}\) (may be seen as part of another expression)
M1: for finding an expression in \(\lambda\) and/or \(\mu\) for one of \(\overrightarrow{KL}\) (or \(\overrightarrow{LK}\)), \(\overrightarrow{LM}\) (or \(\overrightarrow{ML}\)), \(\overrightarrow{KM}\) (or \(\overrightarrow{MK}\))
[If this mark is awarded it assumes the first M1]
M1: for finding an expression in \(\lambda\) or \(\mu\) for two of the following: \(\overrightarrow{KL}\) (or \(\overrightarrow{LK}\)), \(\overrightarrow{LM}\) (or \(\overrightarrow{ML}\)), or \(\overrightarrow{KM}\) (or \(\overrightarrow{MK}\))
A1: dep on M2 for one value correct or both values but written the wrong way round \(\left(\mu = \dfrac{3}{7}\;\lambda = \dfrac{6}{7}\right)\)
A1: dep on M2 for both values
MISREAD \(\overrightarrow{AK} : \overrightarrow{KB} = \lambda : 1\) and \(\overrightarrow{CL} : \overrightarrow{LD} = \mu : 1\)
| Scheme | Marks |
|---|---|
eg \(\overrightarrow{AK} = \left(\dfrac{\lambda}{\lambda + 1}\right)\mathbf{a}\) \(\overrightarrow{KB} = \left(\dfrac{1}{\lambda + 1}\right)\mathbf{a}\) \(\overrightarrow{CL} = \left(\dfrac{-\mu}{1 + \mu}\right)\mathbf{a}\) \(\overrightarrow{DL} = \left(\dfrac{1}{1 + \mu}\right)\mathbf{a}\) or eg \(\overrightarrow{AK} = \left(\dfrac{\frac{1}{2}\mu}{\frac{1}{2}\mu + 1}\right)\mathbf{a} = \left(\dfrac{\mu}{\mu + 2}\right)\mathbf{a}\) \(\overrightarrow{KB} = \left(\dfrac{1}{\frac{1}{2}\mu + 1}\right)\mathbf{a} = \left(\dfrac{2}{\mu + 2}\right)\mathbf{a}\) \(\overrightarrow{CL} = \left(\dfrac{-2\lambda}{1 + 2\lambda}\right)\mathbf{a}\) \(\overrightarrow{DL} = \left(\dfrac{1}{1 + 2\lambda}\right)\mathbf{a}\) | M1 |
eg \(\overrightarrow{KL} = \left(\dfrac{-\lambda}{\lambda + 1}\right)\mathbf{a} + \mathbf{b} + \left(\dfrac{1}{1 + \mu}\right)\mathbf{a}\) or \(\overrightarrow{LM} = \left(\dfrac{-1}{1 + \mu}\right)\mathbf{a} + 0.5\mathbf{b}\) \(\overrightarrow{KM} = \left(\dfrac{-\lambda}{1 + \lambda}\right)\mathbf{a} + \tfrac{3}{2}\mathbf{b}\) oe or eg \(\overrightarrow{KL} = \left(\dfrac{-\frac{1}{2}\mu}{\frac{1}{2}\mu + 1}\right)\mathbf{a} + \mathbf{b} + \left(\dfrac{1}{1 + \mu}\right)\mathbf{a}\) \(\overrightarrow{LM} = \left(\dfrac{-1}{1 + 2\lambda}\right)\mathbf{a} + \tfrac{1}{2}\mathbf{b}\) or \(\overrightarrow{KM} = \left(\dfrac{-\frac{1}{2}\mu}{\frac{1}{2}\mu + 1}\right)\mathbf{a} + \tfrac{3}{2}\mathbf{b}\) oe | M1 |
| Two of the above – may have used \(2\lambda = \mu\) to write all in one of \(\lambda\) or \(\mu\) May be simplified or not – so may have brackets or not | M1 |
| (Giving answers of \(\lambda = 0.5(1 + \sqrt{7})\), \(\mu = 1 + \sqrt{7}\)) |
Notes
M1: For using the ratio to form an expression for a vector passing through \(K\) or \(L\) could be in terms of \(\lambda\) or \(\mu\)
\(\overrightarrow{AK}\) or \(\overrightarrow{KA}\), \(\overrightarrow{KB}\) or \(\overrightarrow{BK}\), \(\overrightarrow{CL}\) or \(\overrightarrow{LC}\), \(\overrightarrow{DL}\) or \(\overrightarrow{LD}\)
(may be seen as part of another expression)
M1: for finding an expression in \(\lambda\) and/or \(\mu\) using the above misread for one of \(\overrightarrow{KL}\) (or \(\overrightarrow{LK}\)), \(\overrightarrow{LM}\) (or \(\overrightarrow{ML}\)), \(\overrightarrow{KM}\) (or \(\overrightarrow{MK}\))
[If this mark is awarded it assumes the first M1]
M1: for finding an expression in \(\lambda\) or \(\mu\) for two of \(\overrightarrow{KL}\) (or \(\overrightarrow{LK}\)), \(\overrightarrow{LM}\) (or \(\overrightarrow{ML}\)), \(\overrightarrow{KM}\) (or \(\overrightarrow{MK}\))
A MAXIMUM OF 3 MARKS CAN BE AWARDED FOR THIS MISREAD