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I was just working on some review textbook problems in James Stewart's Multivariable Calculus when I encountered a problem that looked like the following:

Find the distance between the point $(1,2,3)$ and the $xz$-plane.

I've seen questions that ask for the shortest distance between a point and a plane, but I haven't seen this sort of problem yet.

The way I tackled this problem is by using the distance formula like this:

I called the point $A(1,2,3)$ and called the $xz$-plane as $T(x,0,z)$ and then did $||AT||=\sqrt{(x-1)^2+(-2)^2+(z-3)^2}=\sqrt{(x-1)^2+4+(z-3)^2}$. I'm not sure how to proceed after this. (Maybe this is as far as it goes, because there are infinitely many distances from a point to a plane, depending on what $x$ and $z$ are?)

Any hints would be appreciated. Thanks.

Sujaan Kunalan
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    You are probably reading more into the question than there actually is. I believe the word shortest is implied here. – Mårten W Sep 14 '13 at 20:35

5 Answers5

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In $\mathbf{R}^3$, the distance from the point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is simply $$\frac{|ax_1 + by_1 + cz_1 +d|}{\sqrt{a^2+b^2+c^2}}$$ This is derived in Stewart and is quite easy using the projection of any vector from $(x_1,y_1, z_1)$ to any point in the plane onto a normal vector to the plane. As noted in the comment above, distance usually means shortest distance.

user44441
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  • Could you please post a link for where this formula is proven simply from the vector projections? because I understand those, but struggle to comprehend this ... – Deo Jun 09 '14 at 09:31
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The $xz$ plane is where $y = 0$. In order to get from $(1,2,3)$ to where $y=0$, you have to move $2$ units. So the answer is $2$.

Stefan Smith
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It is just $2 $, here is plot: enter image description here

Ömer
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This is pursuant to user44441's answer; see P800 Example 8, 12.5, Calculus. 6th Ed, by James Stewart. Here's a modified picture to avail:

enter image description here Espy that $\mathbb{n} \neq D$ necessarily.

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$$ {\cal F}\left(\vec{r}\right) = {1 \over 2}\,\left(\vec{r} - \vec{P}\right)^{2} - \mu\,\vec{r}\cdot\hat{y}\,, \quad 0={\partial{\cal F}\left(\vec{r}\right) \over \partial\vec{r}} = \left(\vec{r} - \vec{P}\right) - \mu\,\hat{y} \quad\Longrightarrow\quad \vec{r} = \vec{P} + \mu\,\hat{y} $$

$$ 0 = \vec{r}\cdot\hat{y} = \vec{P}.\cdot\hat{y} + \mu \quad\Longrightarrow\quad \mu = -\vec{P}\cdot\hat{y} \quad\Longrightarrow\quad \vec{r} = \vec{P} - \vec{P}\cdot\hat{y}\ \hat{y} $$

$$ ? = \left\vert\,\vec{r} - \vec{P}\,\right\vert = \left\vert\,\vec{P}\cdot\hat{y}\ \hat{y}\,\right\vert = \left\vert\,P_{y}\,\right\vert = \color{#ff0000}{\Large 2} $$

Felix Marin
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