A perpendicular foot organizes four shapes · source reference

Question

The position vectors of A, B and C relative to origin O are $-\mathbf i+5\mathbf j+\mathbf k$, $\mathbf i-\mathbf j+5\mathbf k$ and $p\mathbf i-4\mathbf j+7\mathbf k$ respectively, where $p$ is constant. Given A, B and C are collinear, show $p=2$. [2]

(i) N divides AC in ratio $AN:NC=\lambda:1-\lambda$. Find $\lambda$ given ON is perpendicular to AC. [4]

(ii) Hence find the exact area of triangle OAC. [2]

(iii) Find the position vector of D such that OADC is a parallelogram. [2]

(iv) Find the coordinates of E such that OAEC is a kite. [2]

[modified 2017/DHS/Promo/6]

Source working

Given $\overrightarrow{OA}=(-1,5,1)$, $\overrightarrow{OB}=(1,-1,5)$, $\overrightarrow{OC}=(p,-4,7)$.

Collinearity implies $\overrightarrow{AB}=k\overrightarrow{AC}$ for some real $k$:

$$(1,-1,5)-(-1,5,1)=k[(p,-4,7)-(-1,5,1)],$$ $$(2,-6,4)=k(p+1,-9,6).$$

Second coordinate: $k=2/3$. First: $2=(2/3)(p+1)\Rightarrow p=3-1=2$ (shown).

(i) Ratio theorem:

$$\begin{aligned}\overrightarrow{ON}&=\lambda\overrightarrow{OC}+(1-\lambda)\overrightarrow{OA}\\&=\lambda(2,-4,7)+(1-\lambda)(-1,5,1)\\&=(-1,5,1)+\lambda(3,-9,6).\end{aligned}$$

$ON\perp AC\Rightarrow\overrightarrow{ON}\cdot\overrightarrow{AC}=0$:

$$[(-1,5,1)+\lambda(3,-9,6)]\cdot[(2,-4,7)-(-1,5,1)]=0,$$ $$[(-1,5,1)+\lambda(3,-9,6)]\cdot(3,-9,6)=0,$$ $$(-3-45+6)+\lambda(9+81+36)=0,$$ $$-42+126\lambda=0\Rightarrow\lambda=1/3.$$

(ii)

$$\overrightarrow{ON}=(-1,5,1)+\tfrac13(3,-9,6)=(0,2,3).$$ $$\begin{aligned}\text{Area}(OAC)&=\tfrac12|\overrightarrow{AC}|\,|\overrightarrow{ON}|\\&=\tfrac12|(3,-9,6)|\,|(0,2,3)|\\&=\tfrac32|(1,-3,2)|\,|(0,2,3)|\\&=\tfrac32\sqrt{1+9+4}\sqrt{4+9}\\&=\tfrac32\sqrt{14}\sqrt{13}=\frac{3\sqrt{182}}2.\end{aligned}$$

(iii) OADC parallelogram implies $\overrightarrow{AD}=\overrightarrow{OC}$, hence $\overrightarrow{OD}-\overrightarrow{OA}=\overrightarrow{OC}$:

$$\overrightarrow{OD}=(-1,5,1)+(2,-4,7)=(1,1,8).$$

(iv) OAEC kite implies $\overrightarrow{OE}=2\overrightarrow{ON}$:

$$\overrightarrow{OE}=2(0,2,3)=(0,4,6).$$

Therefore $E=(0,4,6)$.

Teacher notes

A, B and C collinear means they lie on the same straight line. One may instead consider $\overrightarrow{AB}=k\overrightarrow{BC}$ or $\overrightarrow{BC}=k\overrightarrow{AC}$, for some $k\in\mathbb R$. “Some” refers to a particular value of k.

N is between A and C. N lies outside the segment if the question says N is on AC produced or CA produced.

$\overrightarrow{ON}\cdot\overrightarrow{AC}=0\Rightarrow|ON||AC|\cos\angle ONA=0\Rightarrow\cos\angle ONA=0\Rightarrow\angle ONA=90^\circ$.

Parallelogram vertices are labelled consecutively OADC, clockwise or anticlockwise.

Kite properties: two pairs of adjacent sides are equal, $OA=EA$ and $OC=EC$; diagonals AC and OE meet at right angles; AC bisects OE, so $ON=NE$.

Native diagrams

The source contains a collinear A–B–C diagram; the ratio/height triangle A–N–C with O below and $AN:NC=\lambda:1-\lambda$; its area diagram; parallelogram OADC with N and perpendicular ON; and kite OAEC with equal-side ticks and perpendicular diagonals. These are encoded as verified native geometry.

Printed answer key

(i) $\lambda=1/3$; (ii) $3\sqrt{182}/2$; (iii) $(1,1,8)$; (iv) $(0,4,6)$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 1; solution document pages 1,2. Original page numbers are retained as text metadata.