- Points \(A\) and \(B\) have position vectors \(\mathbf { a }\) and \(\mathbf { b }\), respectively, relative to an origin \(O\), and are such that \(O A B\) is a triangle with \(O A = a\) and \(O B = b\).
The point \(C\), with position vector \(\mathbf { c }\), lies on the line through \(O\) that bisects the angle \(A O B\).
- Prove that the vector \(b \mathbf { a } - a \mathbf { b }\) is perpendicular to \(\mathbf { c }\).
The point \(D\), with position vector \(\mathbf { d }\), lies on the line \(A B\) between \(A\) and \(B\).
- Explain why \(\mathbf { d }\) can be expressed in the form \(\mathbf { d } = ( 1 - \lambda ) \mathbf { a } + \lambda \mathbf { b }\) for some scalar \(\lambda\) with \(0 < \lambda < 1\)
- Given that \(D\) is also on the line \(O C\), find an expression for \(\lambda\) in terms of \(a\) and \(b\) only and hence show that
$$D A : D B = O A : O B$$