# Math Help - Distance between a point and a line Question?

1. ## Distance between a point and a line Question?

Hi everyone, i have a midterm exam coming up and i'm looking at problems in the section of the book to help myself prepare. Can anyone answer this question for me? I have attempted to answer this problem but without the answer i don't know if i'm right.

Find the distance between the point P(1,-2,4) and the line given by the parametric equations:

x= 2t
y= t-3
z= 2t+2

I tried using the techniques given in the book but it's not for a point and line.

Can someone do this one out for me? It's not a homework, i just need to see it done out step by step so i can follow it, along with the answer.

Thank you guys, really desperate...

2. Hello, chrisduluk!

I don't know if you've had Calculus yet.
. . I'll give you a non-Calculus solution.

Find the distance between the point P(1,-2,4) and the line given by:

. . x .= .2t,. . y .= .t - 3,. . z .= .2t + 2
Let point Q be on the line: .Q (2t, t-3, 2t+2)
. . . . . . . . . . . . . . . . . . . . . . . . . . . ____________________________________
The distance from P to Q is: . d . = . √[(2t) - 1]² + [(t - 3) + 2]² + [(2t + 2) - 4]²
. . . . . . . . . . . . . . . . . . . . . . ___________
. . which simplifies to: . d .= .√9t² - 14t + 6 . [1]

And we want to minimize $d.$

If we minimize the square of $d$, we will minimize $d.$

Let $D$ .= . $d^2$ .= .9t² - 14t + 6

This is an up-opening parabola; its minimum point is at its vertex.
The formula for the vertex is: .v .= .-b/2a

. . We have: .a = 9, .b = -14

. . The vertex is: .t .= .-(-14)/(2·9) .= .7/9

. . . . . . . . . . . . . . . . . . . . . _________________ . . . . . ___
Substitute into [1]: . d . = . √9(7/9)² - 14(7/9) + 6 . = . √5/9

. . . . . . . . . . . . . ._
Therefore: . d .= .√5/3

3. Originally Posted by chrisduluk
...

Find the distance between the point P(1,-2,4) and the line given by the parametric equations:

x= 2t
y= t-3
z= 2t+2

...
1. Rewrite the equation of the line:

$(x,y,z) = (0,-3,2) + t \cdot (2,1,2)$

2. Create the equation of a plane E which contains P and has the direction vector of the line as it's normal vector:

$E: (2,1,2) \cdot \left( (x,y,z)-(1, -2, 4)\right)=0~\implies~(2,1,2)\cdot (x,y,z)-8=0$

3. Calculate the coordinates of the point of intersection between the plane E and the given line. Plug in the term of (x, y, z) from the line into the equation of the plane:

$(2,1,2)\cdot \left((0,-3,2) + t \cdot (2,1,2) \right)-8=0 ~\implies~ 1+9t-8 = 0 ~\implies~ t = \frac79$

4. The point of intersection has the coordinates $S\left(\frac{14}9~,~ -\frac{20}9~,~ \frac{32}9 \right)$

5. Calculate the distance between P and S which is the absolute value of $\overrightarrow{PS} = \left(\frac59~,~ -\frac29~,~ -\frac49 \right)$

6. $\left| \overrightarrow{PS} \right| = \sqrt{\frac{25}{81} + \frac{4}{81} + \frac{16}{81}} = \frac19 \sqrt{45} = \frac13 \sqrt{5} \approx 0.745$

4. ## Vector Geometry

We can use vector geometry to represent this problem.
Suppose we have the line through the point $\mathbf{q}$ parallel to the vector $\mathbf{l}$, and a point $\mathbf{p}$ not on the line.
Then $\mathbf{r}=\mathbf{p}-\mathbf{q}$ is the vector from point $\mathbf{q}$ to point $\mathbf{p}$. If it makes an angle θ with the line, and thus with $\mathbf{l}$, then the distance from the point $\mathbf{p}$ to the line is $d=|\mathbf{r}|\sin\theta$. Now, recall that the magnitude of the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is given by
$|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{ b}|\sin\theta$.
Thus,
$d=\frac{|\mathbf{r}\times\mathbf{l}|}{|\mathbf{l}| }$
$d=\frac{|(\mathbf{p}-\mathbf{q})\times\mathbf{l}|}{|\mathbf{l}|}$

Our point is $\mathbf{p}=\mathbf{\hat{i}}-2\mathbf{\hat{j}}+4\mathbf{\hat{k}}$, and our line is that which passes through the point $\mathbf{q}=-3\mathbf{\hat{j}}+2\mathbf{\hat{k}}$ parallel to the vector
$\mathbf{l}=2\mathbf{\hat{i}}+\mathbf{\hat{j}}+2\ma thbf{\hat{k}}$.
Thus
$\mathbf{r}=\mathbf{p}-\mathbf{q}=\mathbf{\hat{i}}+\mathbf{\hat{j}}+2\mat hbf{\hat{k}}$,
and
$d=\frac{|\mathbf{r}\times\mathbf{l}|}{|\mathbf{l}| }$
$=\frac{|(\mathbf{\hat{i}}+\mathbf{\hat{j}}+2\mathb f{\hat{k}})\times(2\mathbf{\hat{i}}+\mathbf{\hat{j }}+2\mathbf{\hat{k}})|}{|2\mathbf{\hat{i}}+\mathbf {\hat{j}}+2\mathbf{\hat{k}}|}$
$=\frac{|2\mathbf{\hat{j}}-\mathbf{\hat{k}}|}{\sqrt{2^2+1^2+2^2}}$
$=\frac{\sqrt{2^2+(-1)^2}}{3}$
$d=\frac{\sqrt{5}}{3}$

--Kevin C.

5. THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote "OH NO" when i said we have to minimize the square of d. Why? Then he gave me 0 points.

6. Originally Posted by Intsecxtanx
THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote "OH NO" when i said we have to minimize the square of d. Why? Then he gave me 0 points.
Only that instructor can really answer that question.
Prehaps he wanted you to use only vector methods.
But as you can see the answers are the same.