A river as the intersection of two planes · source reference

Printed question

Peter is using equations of planes to model two hillsides that meet along a river. The river is modelled by the line where the two planes meet.

One of the hillsides, $H_1$, contains the points $A,B,C$ with coordinates $(3,0,2),(1,0,3),(2,-3,5)$ respectively. The point $A$ is on the river. The other hillside $H_2$ has equation $2x-y+kz=14$, where $k$ is a constant.

(i) Find a vector equation of $H_1$ in scalar product form. [4]

(ii) Show that $k=4$ and deduce that point $B$ is also on the river. [3]

(iii) Write down a cartesian equation of the river. [1]

(iv) Show that $B$ is the point on the river that is nearest to $C$. Hence find the exact shortest distance from $C$ to the river. [3]

(v) Find the acute angle between $BC,H_2$. [2]

[2017/JJC/Prelim/II/3]

Printed working

(i) $\overrightarrow{AB}=(1,0,3)-(3,0,2)=(-2,0,1)$.

$\overrightarrow{AC}=(2,-3,5)-(3,0,2)=(-1,-3,3)$.

A normal to $H_1$ is $\overrightarrow{AB}\times\overrightarrow{AC}=(-2,0,1)\times(-1,-3,3)=(3,5,6)$.

Thus, vector equation of $H_1$ is $\mathbf r\cdot(3,5,6)=(3,0,2)\cdot(3,5,6)$, i.e. $\mathbf r\cdot(3,5,6)=21$.

(ii) $A$ is on the river, the intersection of the 2 planes $\Rightarrow A$ lies on $H_2$.

$\Rightarrow\overrightarrow{OA}\cdot(2,-1,k)=14\Rightarrow(3,0,2)\cdot(2,-1,k)=14\Rightarrow6+2k=14\Rightarrow k=4$ (Shown).

Thus, equation of $H_2$ is $\mathbf r\cdot(2,-1,4)=14$.

Substitute the point $B(1,0,3)$ into equation of $H_2$:

LHS $=(1,0,3)\cdot(2,-1,4)=2+0+12=14=$ RHS.

Since $B$ satisfies the equation of $H_2$, point $B$ lies on $H_2$. Since point $B$ lies on both $H_1,H_2$, thus $B$ lies on the river. (Shown)

(iii) Since $A,B$ lie on the river, a direction vector of river is $\overrightarrow{AB}=(-2,0,1)$.

Thus a vector equation of river is $\mathbf r=(3,0,2)+\lambda(-2,0,1)$.

The cartesian equation of river is $\frac{x-3}{-2}=z-2$, $y=0$.

(iv) $\overrightarrow{BC}=(2,-3,5)-(1,0,3)=(1,-3,2)$.

Since $\overrightarrow{BC}\cdot\overrightarrow{AB}=(1,-3,2)\cdot(-2,0,1)=-2+0+2=0$, $BC$ is perpendicular to $AB$.

$\therefore B$ is the point on the river that is nearest to $C$.

Shortest distance from $C$ to the river $=|\overrightarrow{BC}|=|(1,-3,2)|=\sqrt{1+9+4}=\sqrt{14}$.

(v) Let $\theta$ be acute angle between $BC,H_2$, and $\alpha=90^\circ-\theta$.

Then $\cos\alpha=\frac{|(2,-1,4)\cdot(1,-3,2)|}{|(2,-1,4)||(1,-3,2)|}=\frac{|2+3+8|}{\sqrt{4+1+16}\sqrt{1+9+4}}=\frac{13}{\sqrt{21}\sqrt{14}}$.

$\alpha=40.696^\circ$.

Thus $\theta=90^\circ-40.696^\circ=49.3^\circ$ (correct to 1 d.p.).

Printed answer key

(i) $\mathbf r\cdot(3,5,6)=21$. (iii) $\frac{x-3}{-2}=z-2,y=0$. (iv) $\sqrt{14}$ units. (v) $49.3^\circ$. Short key omits (ii).

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.