1.6: Vectors, collinearity and an area ratio

4PM1/1/November/2020 — Question 11 · 13 marks

1.6 diagram 1
In Figure 5,
OA→=aandOB→=b.\overrightarrow{OA}=\mathbf{a} \qquad \text{and} \qquad \overrightarrow{OB}=\mathbf{b}.
The point C\displaystyle C divides OB\displaystyle OB in the ratio 1:3\displaystyle 1:3. The point D\displaystyle D is the midpoint of AC\displaystyle AC.
(a) Find, as a simplified expression in terms of a\displaystyle \mathbf{a} and b\displaystyle \mathbf{b},
(i) AC→\displaystyle \overrightarrow{AC},
(ii) OD→\displaystyle \overrightarrow{OD},
(iii) BD→\displaystyle \overrightarrow{BD}.
(5)
The point E\displaystyle E is such that OE→=λOA→\displaystyle \overrightarrow{OE}=\lambda\overrightarrow{OA}. Given that E\displaystyle E, D\displaystyle D and B\displaystyle B are collinear,
(b) find the value of λ\displaystyle \lambda.
(4)
Given that
area of △OACarea of △OEB=μ,\frac{\text{area of }\triangle OAC}{\text{area of }\triangle OEB}=\mu,
(c) find the value of μ\displaystyle \mu.
(4)