2013 RVHS P1 Q9

2013 RVHS P1 Q9

10 marks
Prelims

Referred to an origin, \(O\), the position vectors of two points \(A\) and \(B\) are \(\mathbf{a}\) and \(\mathbf{b}\) respectively. \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel.

  1. The point \(C\) lies on \(AB\) produced such that \(AB:BC\) is \(1:3\). Find the position vector of \(C\).[2]
  2. The point \(D\) lies on \(OB\) produced such that \(OB:OD\) is \(1:k\). Given also that \(CD\) is perpendicular to \(AB\), show that \(k=\frac{4{{\left| \mathbf{b} \right|}^{2}}-7\mathbf{a}\cdot \mathbf{b}+3{{\left| \mathbf{a} \right|}^{2}}}{{{\left| \mathbf{b} \right|}^{2}}-\mathbf{a}\cdot \mathbf{b}}\).[3]
  3. Show that \(OADC\) cannot be a parallelogram.[2]
  4. Find the area of triangle \(ABD\) in terms of \(k\).[3]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:(i) \(\mathbf c=4\mathbf b-3\mathbf a\) (ii) \(k=\dfrac{4\lvert\mathbf b\rvert^2-7\mathbf a\cdot\mathbf b+3\lvert\mathbf a\rvert^2}{\lvert\mathbf b\rvert^2-\mathbf a\cdot\mathbf b}\) (iii) \(OADC\) is not a parallelogram (iv) \(\dfrac{k-1}{2}\lvert\mathbf a\times\mathbf b\rvert\) square units

Need help? Join our JC Math tuition classes.

Learn more