Distance between line and a point

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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).

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3 Answers

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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:

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.

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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 $$ \bbox[5px,border:2px solid #C0A000]{\left|\,(p-p_1)-\frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\,\right|}\tag{2} $$

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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 $$ \begin{align} \left|\,(0,-2,2)-\frac{(0,-2,2)\cdot(1,-2,4)}{|(1,-2,4)|^2}(1,-2,4)\,\right| &=\left|\,(0,-2,2)-\frac{12}{21}(1,-2,4)\,\right|\\[6pt] &=\left|\,\frac17(-4,-6,-2)\,\right|\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{2\sqrt{14}}7}\tag{4} \end{align} $$

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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

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

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

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