Referred to the origin $O$, fixed points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf

Question

Referred to the origin $O$, fixed points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where $O$, $A$ and $B$ do not lie on the same line. Point $C$ with position vector $\mathbf{c}$ is a varied point such that $\mathbf{a}+2\mathbf{b}+\lambda \mathbf{c}=\mathbf{0}$, where $\lambda $ is a parameter. Show that $\frac{\mathrm{Area of }\vartriangle ABC}{\mathrm{Area of }\vartriangle ABO}=\left| \frac{\lambda +3}{\lambda } \right|$. [3] Deduce the value of $\lambda $ given that $A$, $B$ and $C$ are collinear. [1] Given instead that $A$, $B$ and $C$ are not collinear and the line $OC$ intersects $AB$ at the point $D$, find the ratio $AD:BD$. [2] Further given that $\angle OAC=\angle OBC={{90}^{{}^\circ }}$, find $\lambda $ in terms of $\left| \mathbf{a} \right|$ and $\left| \mathbf{b} \right|$. [4]

Accepted Answer

(i) $\left|\frac{\lambda+3}{\lambda}\right|$. (ii) $\lambda=-3$. (iii) $AD:BD=2:1$. (iv) $\lambda=\frac{|\mathbf a|^2-4|\mathbf b|^2}{2|\mathbf b|^2-|\mathbf a|^2}$.

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