MHB 231.12.3.65 Determine the smallest distance between a point and a line

[karush]

$\tiny{231.12.3.65}$
$\textsf{Determine the smallest distance between point $P(1,1,1)$ and the line $L$ through the origin $L$ has the direction $\langle -4,-5,8 \rangle$}$

$\textit{ok I'm not real sure I understand this question so assume}\\$
$\textit{the line equation is derived first going thru origin (0,0,0) with direction $\langle -4,-5,8 \rangle$}\\$
$\textit{of which don't know how to do}\\$

$\textit{but then Use the equation }\\$

\begin{align*}\displaystyle
d&=\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}
\end{align*}

$\textit{ Where A,B, and C are coeficients of line equation and $m$, $n$ are coordinates of a point (m,n)}$

 

[karush]

so would the Parametric Form from
$x=x_0+ta\\y=y_0+tb\\z=z_0+tc \\$

be
$L= \, x=-4t, \, y=-5t, \, z=8t$

 

[HallsofIvy]

Any line can be written in the parametric equations of the form x= at+ b, y= ct+ d, z= et+ f. Taking t= 0 at (0, 0, 0) and t= 1 at (-4, -5, 8) we have 0= a(0)+ b, 0= c(0)+ d, z= e(0)+ f, so b= d= f= 0, and -4= a(1)+ b= a, -5= c(1)+ d= c, and 8= e(1)+ f= e. So the equation of the line is x= -4t, y= -5t, and z= 8t.

 

[skeeter]

$\tiny{231.12.3.65}$
$\textsf{Determine the smallest distance between point $P(1,1,1)$ and the line $L$ through the origin $L$ has the direction $\langle -4,-5,8 \rangle$}$

so would the Parametric Form from
$x=x_0+ta\\y=y_0+tb\\z=z_0+tc \\$

be
$L= \, x=-4t, \, y=-5t, \, z=8t$

distance between point $P(1,1,1)$ and any point on line $L$ ...

$d = \sqrt{(-4t-1)^2 + (-5t-1)^2 + (8t-1)^2}$

Shouldn't be that difficult to determine the value of $t$ that minimizes the radicand, $(-4t-1)^2 + (-5t-1)^2 + (8t-1)^2$ ...

 

[karush]

distance between point $P(1,1,1)$ and any point on line $L$ ...

$d = \sqrt{(-4t-1)^2 + (-5t-1)^2 + (8t-1)^2}$

Shouldn't be that difficult to determine the value of $t$ that minimizes the radicand, $(-4t-1)^2 + (-5t-1)^2 + (8t-1)^2$ ...

$\textsf{that turns out to be a parabola}\\$
$\textsf{$(-4t-1)^2 + (-5t-1)^2 + (8t-1)^2 =105 t^2 + 2 t + 3$}\\$

 

[MarkFL]

$\textsf{that turns out to be a parabola}\\$
$\textsf{$(-4t-1)^2 + (-5t-1)^2 + (8t-1)^2 =105 t^2 + 2 t + 3$}\\$

$\textsf{so the vertex is at $(0,\frac{4}{105})$ }$something wacko here

For a parabola of the form:

$$f(x)=ax^2+bx+c$$

We know the axis of symmetry is at:

$$x=-\frac{b}{2a}$$

So, for:

$$f(t)=105t^2+2t+3$$

Where is the axis of symmetry?

Does this parabola open up or down...that is, is the vertex at a minimum or maximum?

 

[karush]

the axis of symmetry is $x=-\frac{1}{105}$

the parabola opens up

$105\left(\frac{1}{105} \right)^2 +2 \left(\frac{1}{105} \right)+3$



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sorry on just a small tablet really hard to use

 

[MarkFL]

the axis of symmetry is $x=-\frac{1}{105}$

the parabola opens up

$105\left(\frac{1}{105} \right)^2 +2 \left(\frac{1}{105} \right)+3$



- - - Updated - - -

sorry on just a small tablet really hard to use

Small quibble...axis of symmetry is:

$$t=-\frac{1}{105}$$

So, since the parabola opens up, the vertex is a minimum, so the minimum distance will be the square root of the value of the parabola on the axis of symmetry. :)

 

[karush]

$$\sqrt{105\left(-\frac{1}{105} \right)^2 -2 \left(\frac{1}{105} \right)+3}=\sqrt{\frac{314}{105}}$$really ?