Question

in the diagram, oabc is a quadrilateral and p, q, r, and s are the mid - points of oa, ab, bc, and oc respectively. \\(\overrightarrow{oa}=\mathbf{a}, \overrightarrow{oc}=\mathbf{b}\\) and \\(\overrightarrow{ab}=\frac{3}{2}\mathbf{a}+\frac{5}{3}\mathbf{b}). find in terms of \\(\mathbf{a}\\) and/or \\(\mathbf{b}) (i) \\(\overrightarrow{ps}), (ii) \\(\overrightarrow{pq}), (iii) \\(\overrightarrow{bc}), (iv) \\(\overrightarrow{sr}). (v) using the information in (ii) and (iv) above or otherwise, state the special name given to quadrilateral pqrs. 8

Explanation:

Step1: Find position vectors of points

Since P is mid - point of OA and $\overrightarrow{OA}=\mathbf{a}$, then $\overrightarrow{OP}=\frac{1}{2}\mathbf{a}$. Since S is mid - point of OC and $\overrightarrow{OC}=\mathbf{b}$, then $\overrightarrow{OS}=\frac{1}{2}\mathbf{b}$.

Step2: Calculate $\overrightarrow{PS}$

$\overrightarrow{PS}=\overrightarrow{OS}-\overrightarrow{OP}=\frac{1}{2}\mathbf{b}-\frac{1}{2}\mathbf{a}=\frac{1}{2}(\mathbf{b}-\mathbf{a})$

Step3: Find position vector of Q

$\overrightarrow{OQ}=\overrightarrow{OA}+\overrightarrow{AQ}$, and since Q is mid - point of AB and $\overrightarrow{AB}=\frac{3}{2}\mathbf{a}+\frac{5}{3}\mathbf{b}$, then $\overrightarrow{AQ}=\frac{1}{2}\overrightarrow{AB}=\frac{3}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$, so $\overrightarrow{OQ}=\mathbf{a}+\frac{3}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}=\frac{7}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$

Step4: Calculate $\overrightarrow{PQ}$

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(\frac{7}{4}\mathbf{a}+\frac{5}{6}\mathbf{b})-\frac{1}{2}\mathbf{a}=\frac{5}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$

Step5: Calculate $\overrightarrow{BC}$

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$, and $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\mathbf{a}+\frac{3}{2}\mathbf{a}+\frac{5}{3}\mathbf{b}=\frac{5}{2}\mathbf{a}+\frac{5}{3}\mathbf{b}$, so $\overrightarrow{BC}=\mathbf{b}-(\frac{5}{2}\mathbf{a}+\frac{5}{3}\mathbf{b})=-\frac{5}{2}\mathbf{a}-\frac{2}{3}\mathbf{b}$

Step6: Find position vector of R

$\overrightarrow{OR}=\overrightarrow{OB}+\overrightarrow{BR}$, and $\overrightarrow{BR}=\frac{1}{2}\overrightarrow{BC}$, so $\overrightarrow{OR}=(\frac{5}{2}\mathbf{a}+\frac{5}{3}\mathbf{b})+\frac{1}{2}(-\frac{5}{2}\mathbf{a}-\frac{2}{3}\mathbf{b})=\frac{5}{2}\mathbf{a}+\frac{5}{3}\mathbf{b}-\frac{5}{4}\mathbf{a}-\frac{1}{3}\mathbf{b}=\frac{5}{4}\mathbf{a}+\frac{4}{3}\mathbf{b}$

Step7: Calculate $\overrightarrow{SR}$

$\overrightarrow{SR}=\overrightarrow{OR}-\overrightarrow{OS}=(\frac{5}{4}\mathbf{a}+\frac{4}{3}\mathbf{b})-\frac{1}{2}\mathbf{b}=\frac{5}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$

Step8: Determine the type of quadrilateral PQRS

Since $\overrightarrow{PQ}=\overrightarrow{SR}=\frac{5}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$, one pair of opposite sides of quadrilateral PQRS is parallel and equal. So PQRS is a parallelogram.

Answer:

(i) $\frac{1}{2}(\mathbf{b}-\mathbf{a})$
(ii) $\frac{5}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$
(iii) $-\frac{5}{2}\mathbf{a}-\frac{2}{3}\mathbf{b}$
(iv) $\frac{5}{4}\mathbf{a}+\frac{5}{6}\mathbf{b}$
(v) Parallelogram