A line that meets the x-axis · source reference

Question

Line l passes through A$(1,2,-1)$ and B$(2,a,1)$. Given l cuts the x-axis, show $a=-2$. [3] Find the shortest distance of O from l. [5]

[modified 2008/NYJC/Promo/5]

Source working

Equation of l:

$$\mathbf r=(1,2,-1)+\lambda(1,a-2,2),\quad\lambda\in\mathbb R.$$

At the x-axis, $y=z=0$.

$$z=0:\quad-1+2\lambda=0\Rightarrow\lambda=1/2,$$ $$y=0:\quad2+\tfrac12(a-2)=0\Rightarrow a=-2.$$

Shortest distance:

$$\begin{aligned}ON&=\frac{|\overrightarrow{OA}\times\overrightarrow{AB}|}{|\overrightarrow{AB}|}\\&=\frac1{\sqrt{21}}|(1,2,-1)\times(1,-4,2)|\\&=\frac1{\sqrt{21}}|(0,-3,-6)|\\&=\frac3{\sqrt{21}}|(0,1,2)|=\frac{3\sqrt5}{\sqrt{21}}\\&=\sqrt{45/21}=\sqrt{15/7}.\end{aligned}$$

Alternative: Find projection length $AN=|\overrightarrow{OA}\cdot\overrightarrow{AB}|/|\overrightarrow{AB}|$. Then Pythagoras gives $ON=\sqrt{OA^2-AN^2}$.

Teacher notes

$l:\mathbf r=\overrightarrow{OA}+\lambda\overrightarrow{AB}$, $\lambda\in\mathbb R$.

The distance formula is

$$\frac{|\overrightarrow{OA}\times\overrightarrow{AB}|}{|AB|}=\frac{|OA||AB|\sin\angle OAB}{|AB|}=OA\sin\angle OAB=OA\sin\theta=ON,$$

the perpendicular distance from O to AB.

Native diagrams

A,N,B are collinear; O is above N; ON is highlighted red and perpendicular to AB; θ at A is $\angle OAB$.

Printed answer key

$\sqrt{15/7}$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 2; solution document pages 8. Original page numbers are retained as text metadata.