A wardrobe model from lines and planes · source reference

Printed question

A home interior designer is looking to design a pull-down wardrobe lift. To test his concept, the designer built a small-scale model of it, in which points $(x,y,z)$ are defined relative to a corner of the model at $O(0,0,0)$. The rods used in the model can be modelled by equations of straight lines.

The main rod can be modelled by the line $l_1$ with equation $\mathbf r=(5,-5,8)+\lambda(1,1,0)$, $\lambda\in\mathbb R$. Another rod, line $l_2$, passes through the points $A(5,-5,8)$ and $B(8,-4,3)$.

(a) A supporting wire connects $B$ to a point $F$ on $l_1$ such that $BF$ is perpendicular to $l_1$. Find the position vector of $F$. [3]

(b) A retractable rod, line $l_R$, is placed such that it can be modelled by reflecting $l_2$ about $l_1$. Find a vector equation of $l_R$. [3]

The panels of the small-scale model can be modelled by equations of planes. The main panel, plane $p_1$, contains the main rod $l_1$ and an attachment point $C(0,0,6)$. A side panel, plane $p_2$, can be represented by the equation $p_2:-17x-37y+4z=24$.

(c) Show that $p_1$ can be represented by the cartesian equation $-x+y+5z=30$. [3]

(d) A structural rod, line $l_S$, needs to be placed at the intersection between $p_1,p_2$. Find a vector equation of $l_S$. [2]

(e) A decorative plank, plane $p_3$, has equation $p_3:-3x+\alpha y+15z=\beta$, where $\alpha,\beta\in\mathbb R$. If the distance between $p_1,p_3$ is exactly $3\sqrt3$ units, find the possible values of $\alpha,\beta$. [3]

[2023/TMJC/Prelim/I/9]

Printed working

(a) $\overrightarrow{OA}=(5,-5,8)$, $\overrightarrow{OB}=(8,-4,3)$.

Method 1: Scalar product = 0

Since $F$ lies on $l_1$, $\overrightarrow{OF}=(5,-5,8)+\lambda(1,1,0)$ for some $\lambda\in\mathbb R$.

Since $BF\perp l_1$, $[(5,-5,8)+\lambda(1,1,0)-(8,-4,3)]\cdot(1,1,0)=0$.

$[(-3+\lambda,-1+\lambda,5)]\cdot(1,1,0)=0\Rightarrow-3+\lambda-1+\lambda=0\Rightarrow\lambda=2$.

$\therefore\overrightarrow{OF}=(5,-5,8)+2(1,1,0)=(7,-3,8)$.

Method 2: Vector Projection

$\overrightarrow{AB}=(8,-4,3)-(5,-5,8)=(3,1,-5)$.

$\overrightarrow{AF}=\frac{\overrightarrow{AB}\cdot(1,1,0)}{|(1,1,0)|}\frac{(1,1,0)}{|(1,1,0)|}=\frac42(1,1,0)=2(1,1,0)=(2,2,0)$.

$\overrightarrow{OF}=\overrightarrow{OA}+\overrightarrow{AF}=(5,-5,8)+(2,2,0)=(7,-3,8)$.

(b) Method 1: Midpoint theorem (w.r.t. point A)

$\overrightarrow{AF}=\frac{\overrightarrow{AB}+\overrightarrow{AB'}}2$, where $B'$ is pt of reflection of $B$ about $l_1$.

$\overrightarrow{AB'}=2\overrightarrow{AF}-\overrightarrow{AB}=2[(7,-3,8)-(5,-5,8)]-[(8,-4,3)-(5,-5,8)]=(1,3,5)$.

$l_R:\mathbf r=(5,-5,8)+\mu(1,3,5)$, $\mu\in\mathbb R$.

Method 2: Midpoint theorem (w.r.t. point O)

$\overrightarrow{OF}=\frac{\overrightarrow{OB}+\overrightarrow{OB'}}2$.

$\overrightarrow{OB'}=2\overrightarrow{OF}-\overrightarrow{OB}=2(7,-3,8)-(8,-4,3)=(6,-2,13)$.

$\overrightarrow{AB'}=(6,-2,13)-(5,-5,8)=(1,3,5)$.

$l_R:\mathbf r=(5,-5,8)+\mu(1,3,5)$, $\mu\in\mathbb R$.

(c) $\overrightarrow{CA}=(5,-5,8)-(0,0,6)=(5,-5,2)$ or $\overrightarrow{CF}=(7,-3,8)-(0,0,6)=(7,-3,2)$.

$\mathbf n_1=(5,-5,2)\times(1,1,0)=((-5)(0)-(2)(1),(2)(1)-(5)(0),(5)(1)-(-5)(1))=(-2,2,10)=2(-1,1,5)$.

$\mathbf r\cdot(-1,1,5)=(0,0,6)\cdot(-1,1,5)=30$.

$p_1:-x+y+5z=30$ (Shown).

(d) Method 1 (using GC)

$p_1:-x+y+5z=30$ — (1); $p_2:-17x-37y+4z=24$ — (2).

Using GC, $x=-21+\frac72t$, $y=9-\frac32t$, $z=t$.

$l_S:\mathbf r=(-21,9,0)+\alpha(7,-3,2)$, $\alpha=\frac t2\in\mathbb R$.

Method 2 (not recommended for this question)

Direction vector of line of intersection

$=(-1,1,5)\times(-17,-37,4)=(189,-81,54)=27(7,-3,2)$.

Check that $C(0,0,6)$ lies on both $p_1,p_2$.

$l_S:\mathbf r=(0,0,6)+\alpha(7,-3,2)$, $\alpha\in\mathbb R$.

(e) $p_1:-x+y+5z=30$; $p_3:-3x+\alpha y+15z=\beta$.

$p_1:\mathbf r\cdot(-1,1,5)=30$; $p_3:\mathbf r\cdot(-3,\alpha,15)=\beta$.

Since $p_1\parallel p_3$, $(-1,1,5)=k(-3,\alpha,15)$, $k=\frac13,\alpha=3$.

Method 1: Length of Projection

Let a point on $p_3$ be $D$, $\overrightarrow{OD}=(-\beta/3,0,0)$.

$\overrightarrow{CD}=(-\beta/3,0,0)-(0,0,6)=(-\beta/3,0,-6)$.

Length of projection $=\frac{|(-\beta/3,0,-6)\cdot(-1,1,5)|}{|(-1,1,5)|}=3\sqrt3$.

$|\frac\beta3-30|=3\sqrt3(\sqrt{27})$.

$\frac\beta3-30=\pm27$.

$\frac\beta3=57$ or $\frac\beta3=3$.

$\beta=171$ or $\beta=9$.

Method 2: Distance between planes

$p_3:\mathbf r\cdot(-3,3,15)=\beta\Rightarrow p_3:\mathbf r\cdot(-1,1,5)=\frac\beta3$ and $p_1:\mathbf r\cdot(-1,1,5)=30$.

Distance between $p_1,p_3$ is exactly $3\sqrt3$ units.

$\frac{|30-\beta/3|}{\sqrt{(-1)^2+1^2+5^2}}=3\sqrt3$.

$|30-\frac\beta3|=27\Rightarrow\frac\beta3=30\pm27\therefore\beta=9$ or $171$.

Teacher notes

For vector projection, the 2 vectors bounding the angle $\theta$ must both be directed away from pt $A$ ($\overrightarrow{AB},\overrightarrow{AF}$) or both be directed towards pt $A$ ($\overrightarrow{BA},\overrightarrow{FA}$).

$p_1\parallel p_3\Rightarrow\mathbf n_1\parallel\mathbf n_3$, where $\mathbf n_1,\mathbf n_3$ are normals of $p_1,p_3$.

Length of projection $=DF=CD\cos\theta=|\overrightarrow{CD}\cdot\hat{\mathbf n}_1|$.

Make the 2 parallel normal vectors of the 2 planes the same, i.e. $(-1,1,5)$.

Note $|(-1,1,5)|=\sqrt{(-1)^2+1^2+5^2}$.

Printed answer key

(a) $(7,-3,8)$. (b) $l_R:\mathbf r=(5,-5,8)+\mu(1,3,5)$, $\mu\in\mathbb R$. (d) $l_S:\mathbf r=(-21,9,0)+\alpha(7,-3,2)$, $\alpha\in\mathbb R$. (e) $\alpha=3,\beta=9$ or $\beta=171$. Short key omits (c).

Editorial qualification

The scalar-product question does not explicitly state parallel planes, but a positive shortest plane-to-plane distance forces parallelism. The source’s normal comparison is valid under this distance interpretation.

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.