Distinguish no intersection from containment · source reference

Printed question

A plane $\pi_1$ has equation $\mathbf r\cdot(2,-1,3)=9$ and a line $l_1$ has equation $\mathbf r=(3,0,0)+\lambda(4,1,-2)$.

(i) Find the coordinates of $P$, the point of intersection of $l_1$ and $\pi_1$. [4]

Hence, or otherwise, find the shortest distance from point $A(3,0,0)$ to $\pi_1$. [2]

The equations of planes $\pi_2,\pi_3$ are given as $\pi_2:\mathbf r=(1,1,1)+s(2,0,-3)+t(2,2,1)$, and $\pi_3:\alpha x+y-z=\beta$, where $\alpha,\beta\in\mathbb R$.

(ii) Find an equation of plane $\pi_2$ in the form $\mathbf r\cdot\mathbf n=d$. Explain why the planes $\pi_1,\pi_2$ intersect. [4]

(iii) The line of intersection of planes $\pi_1,\pi_2$ is $l_2$. The line $l_2$ has equation $\mathbf r=(3,3,2)+\mu(2,1,-1)$. Given that $l_2,\pi_3$ do not have any points in common, find the conditions satisfied by $\alpha,\beta$. [3]

[modified 2010/IJC/Prelim/I/12]

Printed working

(i) $\pi_1:\mathbf r\cdot(2,-1,3)=9$, $l_1:\mathbf r=(3,0,0)+\lambda(4,1,-2)$, $\lambda\in\mathbb R$.

For point of intersection, $[(3,0,0)+\lambda(4,1,-2)]\cdot(2,-1,3)=9$.

$\Rightarrow(3+4\lambda)2-\lambda-6\lambda=9\Rightarrow\lambda=3$.

$\Rightarrow\overrightarrow{OP}=(3,0,0)+3(4,1,-2)=(15,3,-6)$.

Thus, coordinates of $P$ are $(15,3,-6)$.

$\overrightarrow{AP}=(15,3,-6)-(3,0,0)=(12,3,-6)$.

Shortest distance from $A$ to $\pi_1$

$=\frac{|\overrightarrow{AP}\cdot(2,-1,3)|}{|(2,-1,3)|}=\frac1{\sqrt{4+1+9}}|(12,3,-6)\cdot(2,-1,3)|=\frac{|24-3-18|}{\sqrt{14}}=\frac3{\sqrt{14}}=\frac{3\sqrt{14}}{14}$.

Printed alternative solution (otherwise method)

Let $N$ be the foot of perpendicular from point $A$ to the plane $\pi_1$.

Equation of line $AN:\mathbf r=(3,0,0)+\lambda(2,-1,3)$, $\lambda\in\mathbb R$.

At the point of intersection between line $AN$ and the plane,

$[(3,0,0)+\lambda(2,-1,3)]\cdot(2,-1,3)=9$.

$\Rightarrow(3+2\lambda)2+\lambda+9\lambda=9\Rightarrow\lambda=\frac3{14}$.

$\overrightarrow{AN}=\overrightarrow{ON}-\overrightarrow{OA}=(3,0,0)+\frac3{14}(2,-1,3)-(3,0,0)=\frac3{14}(2,-1,3)$.

Thus, the shortest distance $AN=|\overrightarrow{AN}|=\frac3{14}\sqrt{4+1+9}=\frac{3\sqrt{14}}{14}$.

(ii) $\pi_1:\mathbf r\cdot(2,-1,3)=9$, $\pi_2:\mathbf r=(1,1,1)+s(2,0,-3)+t(2,2,1)$.

A normal to plane $\pi_2=(2,0,-3)\times(2,2,1)=(6,-8,4)=2(3,-4,2)$.

$\pi_2:\mathbf r\cdot(3,-4,2)=(1,1,1)\cdot(3,-4,2)=3-4+2=1$, i.e. $\mathbf r\cdot(3,-4,2)=1$.

Since $(3,-4,2)\ne k(2,-1,3)$ for any $k\in\mathbb R$, the normal of planes $\pi_1,\pi_2$ are not parallel to each other. Since the planes are not parallel to each other, they will intersect in a line.

(iii) $l_2:\mathbf r=(3,3,2)+\mu(2,1,-1)$ and $\pi_3:\alpha x+y-z=\beta$, i.e. $\mathbf r\cdot(\alpha,1,-1)=\beta$.

Since $l_2,\pi_3$ do not have any points in common, $l_2$ is parallel to $\pi_3$ and does not lie on $\pi_3$.

So $l_2$ is perpendicular to $(\alpha,1,-1)$ and point $(3,3,2)$ does not lie on $\pi_3$.

Thus $(2,1,-1)\cdot(\alpha,1,-1)=0\Rightarrow2\alpha+1+1=0\Rightarrow\alpha=-1$.

and $(3,3,2)\cdot(\alpha,1,-1)\ne\beta\Rightarrow3(-1)+3-2\ne\beta\Rightarrow\beta\ne-2$.

Teacher notes

Shortest distance from $A$ to $\pi_1=\frac{|\overrightarrow{AP}\cdot\mathbf n|}{|\mathbf n|}=AP\cos\theta=AF$.

For the parametric plane, $(1,1,1)$ is the position vector of a known point on the plane; $(2,0,-3)$ and $(2,2,1)$ are two vectors parallel to the plane.

$l_2\parallel\pi_3\Rightarrow l_2$ is perpendicular to normal to $\pi_3$.

Printed answer key

(i) $(15,3,-6)$; $\frac3{\sqrt{14}}$. (ii) $\mathbf r\cdot(3,-4,2)=1$; normal of $\pi_1,\pi_2$ not parallel → planes not parallel. (iii) $\alpha=-1,\beta\ne-2$.

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.