How to find shortest distance between a point and a vector with projections? (1 Viewer)

SadCeliac · Aug 12, 2023

Hi all,

I don't really have a specific example question from a paper (will look for one), but this is the type of thing that I really can't seem to do - is anyone able to explain this?

I.e., how to use projections properly to find the shortest perpendicular distance between two vectors?

Thanks!

SadCeliac · Aug 12, 2023

E.g.

tywebb · Aug 12, 2023

I wouldn't use projections.

Method 1 (without finding the minimum distance)

$\text{If }d=\text{minimum distance then}$
$d^2=(1+2\lambda)^2+(4+\lambda)^2+(6+\lambda)^2=6\lambda^2+24\lambda+53$
$\text{and }\lambda=\frac{-b}{2a}=\frac{-24}{2\times6}=-2$
$\text{Point is }(1+2\lambda,4+\lambda,6+\lambda)=(1+2(-2),4-2,6-2)=(-3,2,4)$

Method 2 (finding the minimum distance first)

Most efficient way is using cross product.

$\text{Shortest distance from point }{\bf p} \text{ to line }{\bf r}={\bf a}+\lambda{\bf n}\text{ is }d=\frac{|({\bf p}-{\bf a})\times{\bf n}|}{|{\bf n}|}$

$\text{Here }{\bf p}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix},\ {\bf a}=\begin{pmatrix} 1\\ 4\\ 6 \end{pmatrix},\ {\bf n}=\begin{pmatrix} 2\\ 1\\ 1 \end{pmatrix}$

$({\bf p}-{\bf a})\times{\bf n}=\begin{pmatrix} 2\\ -11\\ 7 \end{pmatrix}\text{ so }d=\frac{\sqrt{174}}{\sqrt6}=\sqrt{29}$

$\text{Now }d^2=(1+2\lambda)^2+(4+\lambda)^2+(6+\lambda)^2=29\text{ so }\lambda=-2\text{ and so }$

$\text{point is }(1+2\lambda,4+\lambda,6+\lambda)=(1+2(-2),4-2,6-2)=(-3,2,4)$

SadCeliac · Aug 12, 2023

tywebb said:
I wouldn't use projections either.

Method 1 (without finding the minimum distance)

$\text{If }d=\text{minimum distance then}$
$d^2=(1+2\lambda)^2+(4+\lambda)^2+(6+\lambda)^2=6\lambda^2+24\lambda+53$
$\text{and }\lambda=\frac{-b}{2a}=\frac{-24}{2\times6}=-2$
$\text{Point is }(1+2\lambda,4+\lambda,6+\lambda)=(1+2(-2),4-2,6-2)=(-3,2,4)$

Method 2 (finding the minimum distance first)

Most efficient way is using cross product.

$\text{Shortest distance from point }{\bf p} \text{ to line }{\bf r}={\bf a}+\lambda{\bf n}\text{ is }d=\frac{|({\bf p}-{\bf a})\times{\bf n}|}{|{\bf n}|}$

$\text{Here }{\bf p}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix},\ {\bf a}=\begin{pmatrix} 1\\ 4\\ 6 \end{pmatrix},\ {\bf n}=\begin{pmatrix} 2\\ 1\\ 1 \end{pmatrix}$

$({\bf p}-{\bf a})\times{\bf n}=\begin{pmatrix} 2\\ -11\\ 7 \end{pmatrix}\text{ so }d=\frac{\sqrt{174}}{\sqrt6}=\sqrt{29}$

$\text{Now }d^2=(1+2\lambda)^2+(4+\lambda)^2+(6+\lambda)^2=29\text{ so }\lambda=-2\text{ and so }$

$\text{point is }(1+2\lambda,4+\lambda,6+\lambda)=(1+2(-2),4-2,6-2)=(-3,2,4)$

thank you
only issue is cross product is out of syllabus

for method 1, where did this come from? vvv

tywebb said:
$\text{and }\lambda=\frac{-b}{2a}=\frac{-24}{2\times6}=-2$
$\text{Point is }(1+2\lambda,4+\lambda,6+\lambda)=(1+2(-2),4-2,6-2)=(-3,2,4)$

tywebb · Aug 12, 2023

synthesisFR · Aug 12, 2023

SadCeliac said:
thank you
only issue is cross product is out of syllabus

for method 1, where did this come from? vvv

Vertex of parabola

tywebb · Aug 12, 2023

SadCeliac said:
cross product is out of syllabus

Well if you really must use projections, you can use the formula

$\begin{aligned}d&=|{\bf p}-{\bf a}-\text{Proj}_{\bf n}({\bf p}-{\bf a})|\\&=\textstyle|-{\bf i}-4{\bf j}-6{\bf k}-\frac{(-{\bf i}-4{\bf j}-6{\bf k})\boldsymbol{\cdot}(2{\bf i}+{\bf j}+{\bf k})}{|2{\bf i}+{\bf j}+{\bf k}|^2}(2{\bf i}+{\bf j}+{\bf k})|\\&=\textstyle|-{\bf i}-4{\bf j}-6{\bf k}-\frac{-12}{6}(2{\bf i}+{\bf j}+{\bf k})|\\&=\textstyle|3{\bf i}-2{\bf j}-4{\bf k}|\\&=\sqrt{29}\end{aligned}$

$\text{Now }d^2=(1+2\lambda)^2+(4+\lambda)^2+(6+\lambda)^2=29\text{ so }\lambda=-2\text{ and so }$

$\text{point is }(1+2\lambda,4+\lambda,6+\lambda)=(1+2(-2),4-2,6-2)=(-3,2,4)$

but I think you can now see why I'd rather use cross product.

I guess one argument for projection method is it works in any dimension whereas cross product only works in dimensions 3 and 7.

However in your question it is in 3 dimensions so in this case the cross product method is better.

SadCeliac · Aug 12, 2023

A

tywebb said:
I guess one argument for projection method is it works in any dimension whereas cross product only works in dimensions 3 and 7.

However in your question it is in 3 dimensions so in this case the cross product method is better.

again, cross product is out of syllabus

when I get my trial paper back I'll send the question that I was referring to - it required projections

How to find shortest distance between a point and a vector with projections? (1 Viewer)

SadCeliac

done hsc yay

SadCeliac

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tywebb

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SadCeliac

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tywebb

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synthesisFR

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tywebb

dangerman

SadCeliac

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