1
$\begingroup$

Consider the points (1,2,-1) and (2,0,3).

(a) Find a vector equation of the line through these points in parametric form.

(b) Find the distance between this line and the point (1,0,1). (Hint: Use the parametric form of the equation and the dot product)

I have solved (a), Forming:

Vector equation: (1,2,-1)+t(1,-2,4)

x=1+t

y=2-2t

z=-1+4t

However, I'm a little stumped on how to solve (b).

$\endgroup$
  • $\begingroup$ please refer to en.wikipedia.org/wiki/Distance_from_a_point_to_a_line $\endgroup$ – Anurag A May 27 '15 at 3:33
  • $\begingroup$ @AnuragA I have tried applying the distance formula, but somehow get 0. The answer given is (2/7)(14)^0.5 $\endgroup$ – Larry Lee May 27 '15 at 3:38
  • 1
    $\begingroup$ You need to find the point on the line that is closest to your point--which occurs when the line joining the point to your line is perpendicular to your line. So choose some arbitrary point on the line, find the parametric slope of your point connecting to the arbitrary point (defined by parameter $t$), then take the dot product of that slope and the slope of the original line--set to $0$ and this will give you the value of $t$--then just find the distance between those two points. $\endgroup$ – Jared May 27 '15 at 3:44
  • $\begingroup$ Here's a nonstandard approach I just came up with: Consider a parallelogram, three of whose vertices are $(1,2-1)$, $(2,0,3)$, and $(1,0,1)$. Use the cross product to find the area of that parallelogram, then use "base $\times$ height" to get height -- which is the distance from the line to the point. $\endgroup$ – user137731 May 27 '15 at 4:24
  • $\begingroup$ possible duplicate of Distance of a 3D point from the parametric form of a line? $\endgroup$ – user147263 May 27 '15 at 5:06
5
$\begingroup$

You can use a formula, although I think it's not too difficult to just go through the steps. I would draw a picture first:

enter image description here

You are given that $\vec{p} = (1,0,1)$ and you already found $\vec{m} = (1, -2, 4)$ and $\vec{l}_0 = (1,2,-1)$. Now it's a matter of writing an expression for $\vec{l}(t) - \vec{p}_0$:

\begin{align} \vec{l}(t) - \vec{p}_0 =&\ (\ (t + 1) - 1\ ,\ (-2t + 2) - 0\ ,\ (4t - 1) - 1\ )\\ =&\ (\ t\ ,\ -2t + 2\ ,\ 4t - 2\ ) \end{align}

Now you dot this with the original slope of the line (recall that $\vec{l}(t) - \vec{p}_0$ is the slope of the line segment connecting the point and the line). When this dot product equals zero, you have found $t_0$ and thus $\vec{x}_0$:

\begin{align} \vec{m} \circ (\vec{l}(t) - \vec{p}_0) =&\ (1,-2,4)\circ(\ t\ ,\ -2t + 2\ ,\ 4t - 2\ ) \\ =&\ t + 4t - 4 + 16t - 8 \\ =&\ 21t - 12 \end{align}

Setting this to $0$ gives that $21t_0 - 12 = 0 \rightarrow t_0 = \frac{4}{7}$. This gives the point $\vec{x}_0$ as:

\begin{align} \vec{x}_0 =&\ \vec{l}(t_0) = (\ \frac{4}{7} + 1\ ,\ -\frac{8}{7} + 2\ ,\ \frac{16}{7} - 1\ ) \\ =&\ \frac{1}{7}(11, 6, 9) \end{align}

So finally the distance would be the distance from $\vec{p}_0$ to $\vec{x}_0$:

\begin{align} d =&\ \sqrt{\left(\frac{11}{7} - 1\right)^2 + \left(\frac{6}{7} - 0\right)^2 + \left(\frac{9}{7} - 1\right)^2}\\ =&\ \sqrt{\left(\frac{4}{7}\right)^2 + \left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} \\ =&\ \frac{1}{7}\sqrt{4^2 + 6^2 + 2^2}\\ =&\ \frac{1}{7}\sqrt{56} \\ =&\ \frac{2}{7}\sqrt{14} \end{align}

...or perhaps $\sqrt{\frac{8}{7}}$ is more appealing.

Extra Info

There's no need to worry about whether or not my 2D picture is really representative--it is. No matter how high the dimensions of the problem, the problem itself can always be mapped to exactly 2 dimensions unless the point is on the line--then it's a 1 dimensional problem--which of course we can represent in 2 dimensions just as we can represent this 2 dimensional problem in much higher ones.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

A line through the points $p_1$ and $p_2$ can be written as $$ \bbox[5px,border:2px solid #00A000]{p=p_1+(p_2-p_1)t}\tag{1} $$ The distance from the line in $(1)$ is given by $$ \bbox[5px,border:2px solid #C0A000]{\left|\,(p-p_1)-\frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\,\right|}\tag{2} $$


Plugging in the values for the points $p_1=(1,2,-1)$, $p_2=(2,0,3)$, and $p=(1,0,1)$ into $(1)$ and $(2)$, we get that the line sought is $$ \bbox[5px,border:2px solid #00A000]{(1,2,-1)+(1,-2,4)t}\tag{3} $$ and the distance sought is $$ \begin{align} \left|\,(0,-2,2)-\frac{(0,-2,2)\cdot(1,-2,4)}{|(1,-2,4)|^2}(1,-2,4)\,\right| &=\left|\,(0,-2,2)-\frac{12}{21}(1,-2,4)\,\right|\\[6pt] &=\left|\,\frac17(-4,-6,-2)\,\right|\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{2\sqrt{14}}7}\tag{4} \end{align} $$


Justification of $\boldsymbol{(2)}$

Note that $$ (p-p_1)-\frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\tag{5} $$ and $$ \frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\tag{6} $$ are perpendicular (their dot product is $0$) and sum to $p-p_1$. Thus, they form the triangle

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It's worth noting that the cross product is only defined for three dimensional vectors. $\endgroup$ – Jared May 27 '15 at 5:27
  • 1
    $\begingroup$ @Jared actually, there is a concept of cross product defined on $\mathbb{R}^n$, for $n\ge2$. However, since I have not described that concept, I have edited my answer. $\endgroup$ – robjohn May 27 '15 at 5:34
  • $\begingroup$ I was thinking there should be since (as I just added to my answer) this is after all ultimately a two dimensional problem. $\endgroup$ – Jared May 27 '15 at 5:36
  • $\begingroup$ @Jared: before reading your comment, I had reduced the problem to two dimensions and written it as a formula using the dot product. $\endgroup$ – robjohn May 27 '15 at 5:39
3
$\begingroup$

Using this formula and your computations from (a), we get the expression for distance $d$: $$ d = \frac{\left\| \ \ \left(\ \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} - \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \ \right) \times \, \begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix} \ \ \right\|} { \begin{Vmatrix} 1 \\ -2 \\ 4 \end{Vmatrix} } = \frac{\left\| \ \ \begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix} \times \begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix} \ \ \right\|} { \sqrt{1 + 4 + 16} } = $$

$$ = \frac{1}{\sqrt{21}} \begin{Vmatrix} \vec{\mathbf{i}} & \vec{\mathbf{j}} & \vec{\mathbf{k}} \\ 0 & 2 & -2 \\ 1 & -2 & 4 \end{Vmatrix} = \frac{1}{\sqrt{21}} \begin{Vmatrix} 2 \cdot 4 - (-2) \cdot (-2) \\ (-2)\cdot 1 - 0 \cdot 4 \\ 0 \cdot (-2 ) - 1 \cdot 2 \end{Vmatrix} = \frac{1}{\sqrt{21}} \begin{Vmatrix} 4 \\ -2 \\ -2 \end{Vmatrix} = $$ $$ = \frac{\sqrt{16+4+4}}{\sqrt{21}} = \sqrt{\frac{24}{21}}= \sqrt{\frac{8}{7}} = \frac{2\sqrt{2}}{\sqrt{7}} \approx 1.069 $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The distance between $(1,2,-1)$ and $(1,0,1)$ is way less than $57.6$ and it's very unlikely that $(1,2,-1)$ is the closest point on the line. So you must have made a mistake. $\endgroup$ – user137731 May 27 '15 at 4:21
  • $\begingroup$ @Bye_World You are right, I forgot to compute the unit normal vector for the line. Thank you for the comment. $\endgroup$ – Vlad May 27 '15 at 4:43
  • $\begingroup$ No problem. :-) $\endgroup$ – user137731 May 27 '15 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.