N2015 P1 Q7

N2015 P1 Q7

10 marks
Free

Referred to the origin \(O\), points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively. Point \(C\) lies on \(OA\), between \(O\) and \(A\), such that \(OC : CA = 3 : 2\). Point \(D\) lies on \(OB\), between \(O\) and \(B\), such that \(OD : DB = 5 : 6\).

  1. Find the position vectors \(\overrightarrow{OC}\) and \(\overrightarrow{OD}\), giving your answers in terms of \(\mathbf{a}\) and \(\mathbf{b}\).[2]
  2. Show that the vector equation of the line \(BC\) can be written as \(\mathbf{r} = \frac{3}{5} \lambda \mathbf{a} + (1 - \lambda) \mathbf{b}\), where \(\lambda\) is a parameter. Find in similar form the vector equation of the line \(AD\) in terms of a parameter \(\mu\).[3]
  3. Find, in terms of \(\mathbf{a}\) and \(\mathbf{b}\), the position vector of the point \(E\) where the lines \(BC\) and \(AD\) meet and find the ratio \(AE : ED\).[5]

Video Solution:

Default solution

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Answer:(i) \(\overrightarrow{OC} = \frac{3}{5}\mathbf{a}, \overrightarrow{OD} = \frac{5}{11}\mathbf{b}\) (ii) \(l_{AD} : \mathbf{r} = \frac{5}{11}\mu\mathbf{b} + (1 - \mu)\mathbf{a}, \mu \in \mathbb{R}\) (iii) \(\overrightarrow{OE} = \frac{9}{20}\mathbf{a} + \frac{1}{4}\mathbf{b}, AE : ED = 11 : 9\)

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