Affine collinearity and a triangle area · source reference

Question

Relative to O, A,B,D have position vectors a,b,d. OACB is a parallelogram and $\angle OAC=2\pi/3$ radians.

(i) Given a is a unit vector and $|\mathbf b|=4$, find the length of projection of OC onto OA. [3]

(ii) Given $\lambda\mathbf a+\mu\mathbf b+\mathbf d=\mathbf0$ and $\lambda+\mu+1=0$, show A,B,D are collinear. [3] If $\mu=4$, find area OBD as $k|\mathbf a\times\mathbf b|$, determining k. [3]

[2017/VJC/Prelim/II/2]

Source working

(i)

$$\begin{aligned}\text{projection length}&=|\mathbf c\cdot\hat{\mathbf a}|=|(\mathbf a+\mathbf b)\cdot\hat{\mathbf a}|\\&=|\mathbf a\cdot\hat{\mathbf a}+\mathbf b\cdot\hat{\mathbf a}|=|\mathbf a\cdot\mathbf a+\mathbf b\cdot\mathbf a|\\&=\left||\mathbf a|^2+|\mathbf b||\mathbf a|\cos(\pi-2\pi/3)\right|\\&=|1+4(1/2)|=3.\end{aligned}$$

(ii) Substitute $\lambda=-1-\mu$ into the vector equation:

$$(-1-\mu)\mathbf a+\mu\mathbf b+\mathbf d=\mathbf0\Rightarrow\mu(\mathbf b-\mathbf a)=\mathbf a-\mathbf d,$$ $$\mu\overrightarrow{AB}=\overrightarrow{DA}.$$

AB and DA are parallel and share A, hence A,B,D collinear. At $\mu=4$, $\mathbf d=5\mathbf a-4\mathbf b$.

$$\begin{aligned}\text{Area}(OBD)&=\tfrac12|\mathbf b\times\mathbf d|\\&=\tfrac12|\mathbf b\times(5\mathbf a-4\mathbf b)|\\&=\tfrac12|5\mathbf b\times\mathbf a-4\mathbf b\times\mathbf b|\\&=\tfrac52|\mathbf b\times\mathbf a|=\tfrac52|\mathbf a\times\mathbf b|.\end{aligned}$$

Thus $k=5/2$.

Teacher notes

$\mathbf b\cdot\mathbf a=|\mathbf b||\mathbf a|\cos\angle AOB$.

$\mathbf b\times\mathbf a=-\mathbf a\times\mathbf b$, but $|\mathbf b\times\mathbf a|=|\mathbf a\times\mathbf b|$. Also $\mathbf b\times\mathbf b=\mathbf0$.

Native diagrams

Parallelogram O,A,C,B in consecutive order, with diagonal OC and interior angle $2\pi/3$ at A.

Printed answer key

(i) $3$; (ii) $k=5/2$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 2; solution document pages 11. Original page numbers are retained as text metadata.