Parallel lines, a parallelogram and a midplane · source reference

Printed question

The points $A,B$ have position vectors $3\mathbf i+\mathbf j$ and $3\mathbf i+3\mathbf j$ respectively. The line $l_1$ and the planes $\Pi_1,\Pi_2$ have equations as follows:

$l_1:\mathbf r=\overrightarrow{OA}+\alpha(2,1,-1)$, $\Pi_1:x+2z=3$, $\Pi_2:\mathbf r=\lambda(1,1,0)+\mu(0,1,1)$, where $\alpha,\lambda,\mu\in\mathbb R$.

It is given that the planes $\Pi_1,\Pi_2$ intersect in the line $l_2$.

(i) Find a vector equation of the line $l_2$ and show that the line $l_2$ is parallel to the line $l_1$. Hence find the shortest distance between the lines $l_1,l_2$.

(ii) The points $C,D$ are on the lines $l_1,l_2$ respectively such that $\angle ABC=90^\circ$ and $ABDC$ is a parallelogram. Find the position vectors of $C,D$. Hence find the area of the parallelogram $ABDC$.

(iii) The plane $\Pi_3$ is parallel to the plane $\Pi_2$ and is equidistant to both point $A$ and the plane $\Pi_2$. Show that the equation of the plane $\Pi_3$ is given by $\mathbf r\cdot(\mathbf i-\mathbf j+\mathbf k)=1$.

(iv) Find the position vector of the foot of the perpendicular from the point $A$ to the plane $\Pi_3$.

Source wording note: In the first sentence of part (iii), the question PDF prints $\Pi_2$ where the subsequent reference to $\Pi_3$ identifies the intended plane. This transcription uses $\Pi_3$ there; the original wording is recorded as a source typo.

Printed working

(i) $\Pi_1:x+2z=3\Rightarrow\mathbf r\cdot(1,0,2)=3$; $\Pi_2:\mathbf r=\lambda(1,1,0)+\mu(0,1,1)$, $\lambda,\mu\in\mathbb R$.

To find $l_2$, the line of intersection of $\Pi_1,\Pi_2$, consider

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

Equation of line $l_2:\mathbf r=(3-2\mu)(1,1,0)+\mu(0,1,1)$, i.e. $\mathbf r=(3,3,0)+\mu(-2,-1,1)$, $\mu\in\mathbb R$.

Since the direction vector of $l_1=(2,1,-1)=-$ direction vector of $l_2$, $l_2\parallel l_1$ (Shown).

Since point $A$ lies on $l_1$, point $B$ lies on $l_2$, shortest distance

$=\frac{|\overrightarrow{AB}\times(2,1,-1)|}{|(2,1,-1)|}=\frac1{\sqrt6}|(0,2,0)\times(2,1,-1)|=\frac1{\sqrt6}|(-2,0,-4)|=\frac{\sqrt{30}}3$.

(ii) $l_1:\mathbf r=(3,1,0)+\alpha(2,1,-1)$.

Since $C$ lies on $l_1$, $\overrightarrow{OC}=(3,1,0)+\alpha(2,1,-1)$ for some $\alpha\in\mathbb R$.

$\overrightarrow{BA}=(0,-2,0)$ and $\overrightarrow{BC}=(3,1,0)+\alpha(2,1,-1)-(3,3,0)=(0,-2,0)+\alpha(2,1,-1)$.

Since $BA\perp BC$, $\overrightarrow{BA}\cdot\overrightarrow{BC}=0$.

$\Rightarrow[(0,-2,0)+\alpha(2,1,-1)]\cdot(0,-2,0)=0\Rightarrow4-2\alpha=0\Rightarrow\alpha=2$.

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

Since $ABCD$ is a parallelogram, $\overrightarrow{CD}=\overrightarrow{AB}$.

$\Rightarrow\overrightarrow{OD}-(7,3,-2)=(0,2,0)\Rightarrow\overrightarrow{OD}=(7,5,-2)$.

Area of parallelogram $ABDC=|\overrightarrow{BA}\times\overrightarrow{BC}|$

$=|(0,-2,0)\times(4,0,-2)|=4|(0,-1,0)\times(2,0,-1)|=4|(1,0,2)|=4\sqrt5$.

(iii) $\Pi_2:\mathbf r=\lambda(1,1,0)+\mu(0,1,1)$.

A normal to $\Pi_2=(1,1,0)\times(0,1,1)=(1,-1,1)$.

Note that origin $O$ lies on $\Pi_2$. Let $M$ be the mid-point of $OA$.

Then $\overrightarrow{OM}=\frac12\overrightarrow{OA}=\frac12(3,1,0)$.

Since $\Pi_3$ is parallel to $\Pi_2$ and is equidistant to both point $A$ and plane $\Pi_2$, $M$ lies on $\Pi_3$, and $(1,-1,1)$ is a normal of $\Pi_3$.

Thus, equation of $\Pi_3$ is $\mathbf r\cdot(1,-1,1)=\frac12(3,1,0)\cdot(1,-1,1)\Rightarrow\mathbf r\cdot(1,-1,1)=1$.

(iv) Let $F$ be foot of perpendicular from $A$ to $\Pi_3$.

Line $AF:\mathbf r=(3,1,0)+\delta(1,-1,1)$, $\delta\in\mathbb R$.

$\Pi_3:\mathbf r\cdot(1,-1,1)=1$.

To find point of intersection between line $AF$ and plane $\Pi_3$, consider

$[(3,1,0)+\delta(1,-1,1)]\cdot(1,-1,1)=1$.

$\Rightarrow(3-1)+\delta(1+1+1)=1\Rightarrow\delta=-\frac13$.

Hence, $\overrightarrow{OF}=(3,1,0)-\frac13(1,-1,1)=\frac13(9-1,3+1,0-1)=\frac13(8,4,-1)$.

Teacher notes

Shortest distance $=BN=AB\sin\theta=|\overrightarrow{AB}\times\hat{\mathbf d}|$, where $\mathbf d=(2,1,-1)$.

Printed answer key

(i) $\mathbf r=(3,3,0)+\mu(-2,-1,1)$, $\mu\in\mathbb R$; $\frac{\sqrt{30}}3$. (ii) $\overrightarrow{OC}=(7,3,-2)$; $\overrightarrow{OD}=(7,5,-2)$; $4\sqrt5$ units$^2$. (iv) $\frac13(8,4,-1)$. Short key omits (iii).

Editorial qualification

The printed continuation says “ABCD is a parallelogram”, whereas the question specifies ABDC. The used equation CD=AB and computed D=(7,5,−2) belong to ABDC. The source lettering typo is retained. The area formula uses BA×BC: this equals the area of ABDC because BD=BA+AC and BA×BC differs only by sign from AB×AC.

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.