Equal radii make a midpoint perpendicular · source reference

Question

A and B have position vectors $\mathbf a,\mathbf b$ relative to O, with $|\mathbf a|=|\mathbf b|$. P with position vector $\mathbf p$ lies on AB such that $\mathbf b\cdot\mathbf p=\mathbf a\cdot\mathbf p$.

(i) Show AB is perpendicular to OP. [2]

(ii) Determine the position vector of D in terms of a,b, where D is O reflected about AB. [2]

(iii) Give the geometrical meaning of $|\mathbf a\times\mathbf b|$. [1]

[2010/NJC/Prelim/I/3]

Source working

(i)

$$\begin{aligned}\overrightarrow{AB}\cdot\overrightarrow{OP}&=(\mathbf b-\mathbf a)\cdot\mathbf p\\&=\mathbf b\cdot\mathbf p-\mathbf a\cdot\mathbf p\\&=\mathbf a\cdot\mathbf p-\mathbf a\cdot\mathbf p=0.\end{aligned}$$

Hence AB is perpendicular to OP.

Alternative: $\mathbf b\cdot\mathbf p=\mathbf a\cdot\mathbf p\Rightarrow(\mathbf b-\mathbf a)\cdot\mathbf p=0\Rightarrow\overrightarrow{AB}\cdot\overrightarrow{OP}=0$.

(ii) Since $|\mathbf a|=|\mathbf b|$, P is the midpoint of AB. Ratio theorem gives $\overrightarrow{OP}=\tfrac12(\mathbf a+\mathbf b)$.

$$\overrightarrow{OD}=2\overrightarrow{OP}=2[\tfrac12(\mathbf a+\mathbf b)]=\mathbf a+\mathbf b.$$

(iii) $|\mathbf a\times\mathbf b|$ represents the area of rhombus OADB or OBDA.

Teacher notes

The margin derives

$$|\mathbf a\times\mathbf b|=(OA)(OB)\sin\angle AOB=2\times\tfrac12(OB)(h)=2\times\text{Area}(OAB)=\text{Area of rhombus OADB}.$$

Native diagrams

The rhombus has O above, D below, A left, B right and P at the perpendicular diagonal intersection. All four side-equality ticks, angle bisector at O, right angle at P, altitude h from A to OB and its right-angle mark are retained. Degenerate A=B would not define a unique line AB; the printed geometric construction presumes distinct A,B.

Printed answer key

(ii) $\mathbf a+\mathbf b$; (iii) area of rhombus OADB or OBDA.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 1; solution document pages 6,7. Original page numbers are retained as text metadata.