12.3.63 Determine the smallest distance between point and a line

[karush]

$\text{Determine the smallest distance between point}$
$$P(1,1,1)$$
$\textsf{ and the line $L$ through the origin $L$ has the direction}$
$$\langle -4,-5,8 \rangle$$

ok just barely had time to post this
but was ?? about direction

presume going off of this

$\textit{Distance between point and line }$
$\textit{ Where A,B, and C are coeficients of line equation And M, N are coordinates of a point}$
\begin{align*}\displaystyle
d&=\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}
\end{align*}

 

[MarkFL]

A vector along the line is given by:

\(\displaystyle \vec{v}=\left[\begin{array}{c}-4t \\ -5t \\ 8t \end{array}\right]\)

So, what is the square of the distance between a point on the line with parameter $t$ and the point $(1,1,1)$?

 

[karush]

A vector along the line is given by:

\(\displaystyle \vec{v}=\left[\begin{array}{c}-4t \\ -5t \\ 8t \end{array}\right]\)

So, what is the square of the distance between a point on the line with parameter $t$ and the point $(1,1,1)$?

ok not real sure what you mean by this

$\sqrt{(-4)^2+(-5)^2+(8)^2}$ ?

 

[MarkFL]

ok not real sure what you mean by this

$\sqrt{(-4)^2+(-5)^2+(8)^2}$ ?

Let's back up a bit...a point on the line is:

\(\displaystyle (-4t,-5t,8t)\)

And the given point is:

\(\displaystyle (1,1,1)\)

What is the square of the distance between these two points?

 

[karush]

sorry this looks like duplicate post done much earlier
somehow it never got recorded on the HW

mod can delete :(

 

[MarkFL]

sorry this looks like duplicate post done much earlier
somehow it never got recorded on the HW

mod can delete :(

Yes, this is a duplicate of the thread posted here:

http://mathhelpboards.com/calculus-10/231-12-3-65-determine-smallest-distance-between-point-line-22060.html

The method used there was what I had in mind for finding the required distance.

However, let's take this opportunity to develop a general formula. Suppose we have a line described by the two points:

\(\displaystyle P_1\left(x_1,y_1,z_1\right),\,P_2\left(x_2,y_2,z_2\right)\)

And the point:

\(\displaystyle P_0\left(x_0,y_0,z_0\right)\)

And so a vector along the line can be given as:

\(\displaystyle \left[\begin{array}{c}x_1+\left(x_2-x_1\right)t \\ y_1+\left(y_2-y_1\right)t \\ z_1+\left(z_2-z_1\right)t \end{array}\right]\)

And thus, the square of the distance $D$ between a point on the line with parameter $t$ and the given point is:

\(\displaystyle f(t)=D^2(t)=\left(\left(x_1-x_0\right)+\left(x_2-x_1\right)t\right)^2+\left(\left(y_1-y_0\right)+\left(y_2-y_1\right)t\right)^2+\left(\left(z_1-z_0\right)+\left(z_2-z_1\right)t\right)^2\)

Expanding, we get:

\(\displaystyle f(t)=\left(\left(x_1-x_0\right)^2+\left(y_1-y_0\right)^2+\left(z_1-z_0\right)^2\right)+2\left(\left(x_1-x_0\right)\left(x_2-x_1\right)+\left(y_1-y_0\right)\left(y_2-y_1\right)+\left(z_1-z_0\right)\left(z_2-z_1\right)\right)t+\left(\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right)t^2\)

Using vector notation, this becomes:

\(\displaystyle f(t)=\left|\vec{P_1}-\vec{P_0}\right|^2+2\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)t+\left|\vec{P_2}-\vec{P_1}\right|^2t^2\)

Differentiating w.r.t $t$ and equating the result to 0 and solving for $t$, we find:

\(\displaystyle t=-\frac{\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)}{\left|\vec{P_2}-\vec{P_1}\right|^2}\)

Observing that $f$ is a parabolic function opening upwards, we can then state:

\(\displaystyle f_{\min}=f\left(-\frac{\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)}{\left|\vec{P_2}-\vec{P_1}\right|^2}\right)=\left|\vec{P_1}-\vec{P_0}\right|^2-\frac{\left(\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)\right)^2}{\left|\vec{P_2}-\vec{P_1}\right|^2}\)

\(\displaystyle f_{\min}=\frac{\left|\vec{P_1}-\vec{P_0}\right|^2\left|\vec{P_2}-\vec{P_1}\right|^2-\left(\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)\right)^2}{\left|\vec{P_2}-\vec{P_1}\right|^2}\)

Using the vector quadruple product, we may write:

\(\displaystyle f_{\min}=\left(\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}\right)^2\)

Hence:

\(\displaystyle D_{\min}=\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}\)

Now, in the given problem, we may take:

\(\displaystyle \vec{P_0}=\langle 1,1,1 \rangle\)

\(\displaystyle \vec{P_1}=\langle 0,0,0 \rangle\)

\(\displaystyle \vec{P_2}=\langle -4,-5,8 \rangle\)

And so we have:

\(\displaystyle \vec{P_2}-\vec{P_1}=\langle -4,-5,8 \rangle\)

\(\displaystyle \vec{P_1}-\vec{P_0}=\langle -1,-1,-1 \rangle\)

And we then find:

\(\displaystyle \left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)=\langle 13,-12,-1 \rangle\)

So, we have:

\(\displaystyle D_{\min}=\sqrt{\frac{13^2+12^2+1^2}{4^2+5^2+8^2}}=\sqrt{\frac{314}{105}}\quad\checkmark\)

 

[karush]

always surprised how many views
these get