In triangle $OAB$, $\angle OAB=90{}^\circ $ and the point $C$ on $AB$ is such that $AC=\frac{2}{3}CB

Question

In triangle $OAB$, $\angle OAB=90{}^\circ $ and the point $C$ on $AB$ is such that $AC=\frac{2}{3}CB$. With respect to the origin $O$, the position vectors of $A$ and $B$ are given as $\mathbf{a}$ and $\mathbf{b}$ respectively. Show that $\mathbf{a}\cdot \mathbf{b}={{\left| \mathbf{a} \right|}^{2}}$.[1] Find $\mathbf{c}$, the position vector of $C$ in terms of $\mathbf{a}$ and $\mathbf{b}$.[1] Given that the lengths of $OA$ and $OB$ are $3$ and $5$ units respectively, find the length of projection of $\mathbf{c}$ onto $\mathbf{b}$.[3]

Accepted Answer

(i) $\mathbf a\cdot\mathbf b=\lvert\mathbf a\rvert^2$ (ii) $\mathbf c=\dfrac{3\mathbf a+2\mathbf b}{5}$ (iii) $\dfrac{77}{25}$ units

2008 IJC P1 Q2

2008 IJC P1 Q2

Video Solution:

Solution:

Solution locked