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

Question

Referred to the origin $O$, points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. It is also given that $OAB$ is a straight line (see diagram). Show that the area of triangle $ABC$ can be written in the form $k\left| \left( \mathbf{b}-\mathbf{a} ight)	imes \mathbf{c} ight|$, where $k$ is a constant to be determined. [3] It is given that $\overrightarrow{AB}$ is a unit vector and $C$ is equidistant from $A$ and $B$. Give a geometrical interpretation of $\left| \left( \mathbf{a}-\mathbf{b} ight)	imes \mathbf{c} ight|$. [1] Show that $OB$ has length $\left| \mathbf{c}\cdot \mathbf{b}-\mathbf{c}\cdot \mathbf{a} ight|+q$, where $q$ is a constant to be determined. [3]

Accepted Answer

(i) k = 1/2 (shown); (ii) Shortest distance from C to line AB, or perpendicular height of △ABC; (iii) q = 1/2

2022 VJC P2 Q1

2022 VJC P2 Q1

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