2023 TJC Promo Q6

2023 TJC Promo Q6

8 marks
Promo

With reference to the origin \(O\), the points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively where \(\mathbf{a}\) and \(\mathbf{b}\) are non-zero and non-parallel vectors. Point \(C\) lies between \(O\) and \(A\) such that \(OC:CA=3:2\). Point \(X\) lies between \(A\) and \(B\) such that \(AX:XB=1:3\). The lines \(BC\) and \(OX\) meet at the point \(E\).

  1. Show that the vector equation of the line \(BC\) can be written as \(\mathbf{r}=\frac{3}{5}\lambda \mathbf{a}+\left( 1-\lambda \right)\mathbf{b}\) where \(\lambda \) is a parameter.[1]
  2. By finding a vector equation of the line \(OX\) and using the result in (a), show that the position vector of \(E\) is \(\frac{1}{2}\mathbf{a}+\frac{1}{6}\mathbf{b}\).[4]
  3. It is given that angle \(AOB\) is \({{30}^{\circ }}\) and \(\mathbf{b}\) is a unit vector. Find the exact perpendicular distance from \(E\) to the line \(OA\).[3]

Video Solution:

Video Solution

Video solution locked

Solution:

Solution locked

Sign in to view the step-by-step solution

Similar questions are unavailable for this question.
Answer:(a) \(\mathbf r=\dfrac35\lambda\mathbf a+(1-\lambda)\mathbf b\), \(\lambda\in\mathbb R\) (b) \(\overrightarrow{OE}=\dfrac12\mathbf a+\dfrac16\mathbf b\) (c) \(\dfrac1{12}\text{ unit}\)

Need help? Join our JC Math tuition classes.

Learn more