The shortest cable to an inclined roof diagonal · source reference

Printed question

The diagram above shows a partial design of the roof of the gallery of an amphitheatre. $ABCD$ is an inclined rectangular roof, where $AB=10$ m and $BC=5$ m. $EFGH$ is a rectangle on a horizontal ground, where $EF=10$ m and $EH=4$ m. The points $A,B$ are 3 m directly above $E,F$ respectively. The points $C,D$ are 6 m directly above $G,H$ respectively.

Point $O$ is the centre of $EFGH$ and is taken as the origin. Perpendicular unit vectors $\mathbf i,\mathbf j,\mathbf k$ are in the direction of $\overrightarrow{EF},\overrightarrow{EH},\overrightarrow{EA}$ respectively.

(i) Find a vector equation of the line passing through $A,C$. [2]

(ii) A spot light is fixed at the top of a vertical pillar erected at point $O$ with coordinates $(15,6,0)$. Find the height of the pillar if the top of the pillar is collinear with $A,C$. [3]

(iii) The main electrical supply to the gallery runs along the diagonal $AC$. Find the position vector of point $X$ on $AC$ such that the length of the electric cable that is required to provide electricity from the main electrical supply to an amplifier mounted at $D$, will be the least. [3]

[2010/PJC/Prelim/II/3]

Printed working

(i) $\overrightarrow{AC}=(5,2,6)-(-5,-2,3)=(10,4,3)$.

Line $AC:\mathbf r=(5,2,6)+\lambda(10,4,3)$, $\lambda\in\mathbb R$, or $\mathbf r=(-5,-2,3)+\lambda(10,4,3)$, $\lambda\in\mathbb R$.

(ii) Let point $R$ be the top of pillar of height $h$, i.e. $\overrightarrow{OR}=(15,6,h)$.

Since $R$ lies on line $AC$, $\overrightarrow{OR}=(5,2,6)+\lambda(10,4,3)$ for some real $\lambda$.

Equating $(15,6,h)=(5,2,6)+\lambda(10,4,3)$.

$15=5+10\lambda\Rightarrow\lambda=1\Rightarrow h=6+3\lambda=9$.

Thus, height of pillar is 9 m.

Alternative solution

Since $A,C,R$ are collinear points, $\overrightarrow{AC}=k\overrightarrow{AR}$ for some real $k$.

$\Rightarrow(10,4,3)=k[(15,6,h)-(-5,-2,3)]=k(20,8,h-3)$.

From 1st row: $10=20k\Rightarrow k=\frac12$.

From 3rd row: $3=\frac12(h-3)\Rightarrow h=9$.

Thus, height of pillar is 9 m.

(iii) Since $X$ lies on $AC$, $\overrightarrow{OX}=(5,2,6)+\lambda(10,4,3)$ for some real $\lambda$.

For $DX$ to be the least, $DX\perp AC$, i.e. $\overrightarrow{DX}\cdot\overrightarrow{AC}=0$.

$\overrightarrow{DX}=(5,2,6)+\lambda(10,4,3)-(-5,2,6)=(10,0,0)+\lambda(10,4,3)$.

$[(10,0,0)+\lambda(10,4,3)]\cdot(10,4,3)=0$.

$\Rightarrow100+\lambda(100+16+9)=0\Rightarrow\lambda=-\frac45$.

Thus $\overrightarrow{OX}=(5,2,6)-\frac45(10,4,3)=\frac35(-5,-2,6)$.

Printed answer key

(i) $\mathbf r=(5,2,6)+\lambda(10,4,3)$, $\lambda\in\mathbb R$, or $\mathbf r=(-5,-2,3)+\lambda(10,4,3)$, $\lambda\in\mathbb R$. (ii) 9 m. (iii) $\frac35(-5,-2,6)$.

Editorial qualification

The question first defines O as the origin, then calls the pillar base O with coordinates (15,6,0). These are incompatible point names. The printed working uses a pillar top R=(15,6,h); Porisma retains the original wording and interprets the base as a separately named point (15,6,0).

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question booklet PDF pages 4–12 (printed 80–88); solution booklet PDF pages 17–44. Column vectors are typeset as ordered coordinate triples here; the entries, products, intermediate equations and original parameter/domain restrictions are retained.