$ax+by+cz=d$ is the equation of a plane in space. Show that $d'$ is the distance from the plane to the origin.

Question

This is a 3 part practice question I would like to get some feedback on. I think I have solved the 1st two parts, but I need a little direction for part (c) (the title is Part (a) ) which is repeated here with more detail,

a) $ax+by+cz=d$ is the equation of a plane in space. Divide this equation by a real number to get $a'x+b'y+c'z=d'$.

Show that $d'$ is the distance from the plane to the origin.

It is required that $ a'^2 + b'^2 + c'^2 = 1 $ and $d' >= 0$

Here is my solution to part (a)

Let $p$ be a real number $ (p \neq 0)$, such that

$\frac{1}{p}(ax + by + cz) = \frac{d}{p} = a'x + b'y + c'z = d'$

It is required that

$ a'^2 + b'^2 + c'^2 = 1 $ so,

$ (\frac{a}{p})^2 + (\frac{b}{p})^2 + (\frac{c}{p})^2 = 1 $, this implies

$ a^2 + b^2 + c^2 = p^2 $

$ \pm\sqrt{a^2 + b^2 + c^2} = p $

Since $d = ax_0 + by_0 + cz_0 $ (where $ Q(x_0,y_0,z_0) $ is a point on the plane),

$d' = \frac{1}{p}(ax_0 + by_0 + cz_0)$

Substituting $ p $ into $d'$, we get

$d' = \frac{1}{\sqrt{a^2 + b^2 + c^2}}(ax_0 + by_0 + cz_0) = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}}$

(taking the positive since it is required that $d' >= o$)

Since

$ ai + bj + ck $ is a normal vector to the plane and,

$ x_0i + y_0j + z_0k $ is the position vector of point $Q$

$d' = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}} = \frac{|n \cdot \vec{\mathbf{OQ}|} }{|n|}$

This is the formula from a point to a plane (in this case the origin).

b) Find the position vector of the point $S$ that is closest to the origin on the plane, with equation in the primed form of part (a).

My solution to part (b):

From part (a) it was determined that

$d' = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}} = \frac{|n \cdot \vec{\mathbf{OQ}|} }{|n|}$ , is the distance from the plane to the origin.

And, in the primed form

$ \vec{\mathbf{n}} = a'i + b'j + c'k $ is a normal vector to the plane

Since $ \vec{\mathbf{n}} \cdot \vec{\mathbf{n}} = |\vec{\mathbf{n}}|^2 = a'^2 + b'^2 + c'^2 = 1 $ so $ |\vec{\mathbf{n}}| = 1$

$ \vec{\mathbf{n}} $ is the unit normal vector to $ a'x + b'y + c'z = d'$

The position vector $ \vec{\mathbf{S}} $ is then:

$ d'\vec{\mathbf{n}} = d' (a'i + b'j + c'k) $

Is this a sufficient answer? Or should I expand d'?

c) Consider the plane given by the equation in the primed form of part (b). Why is each plane containing the origin described by two distinct equations of this form? Why does each plane that does not contain the origin have a unique equation of this form?

This question, I need a little direction.

If the equation of a plane in form $ a'x + b'y + c'z = d'$ contains the origin, then

$d' = 0$, so

$ a'x + b'y + c'z = \frac{a}{p}x + \frac{b}{p}x + \frac{c}{p}x = 0$

What would be a 2nd distinct equation of this form? Could someone point me in the right direction?

Answer 1

Your answer looks good. Here is an alternate method to find the minimum point.

To minimize $$ x^2+y^2+z^2\tag{1} $$ while maintining $$ ax+by+cz=d\tag{2} $$ we want to have $$ 2x\,\delta x+2y\,\delta y+2z\,\delta z=0\tag{3} $$ for all $(\delta x,\delta y,\delta z)$ so that $$ a\,\delta x+b\,\delta y+c\,\delta z=0\tag{4} $$ By orthogonality, this means we need $(a,b,c)\|(x,y,z)$; that is, $$ (x,y,z)=k(a,b,c)\tag{5} $$ Substituting $(5)$ into $(2)$ yields $$ k(a^2+b^2+c^2)=d\tag{6} $$ Plugging $(5)$ and $(6)$ into $(1)$, we get the square of the minimum distance is $$ \left(\frac{d}{a^2+b^2+c^2}\right)^2(a^2+b^2+c^2)\tag{7} $$ Thus, the minimum distance is $$ \frac{|d|}{\sqrt{a^2+b^2+c^2}}\tag{8} $$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ {\cal F}\equiv\half\,x^{2} + \half\,y^{2} + \half\,z^{2} - \mu\pars{ax + by + cz - d} $$

$$ x=\mu a\,,\quad y=\mu b\,,\quad z=\mu c $$

$$ a\,\pars{\mu a} + b\,\pars{\mu b} + c\,\pars{\mu c} = d\quad\imp\quad \mu = {d \over a^{2} + b^{2} + c^{2}} $$

\begin{align} \color{#00f}{\large\mbox{Distance}}&= \root{x^{2} + y^{2} + z^{2}} =\verts{\mu}\root{a^{2} + b^{2} + c^{2}} ={\verts{d} \over \root{a^{2} + b^{2} + c^{2}}} \\[3mm]&=\color{#00f}{\large% {1 \over \root{\pars{a/d}^{2} + \pars{b/d}^{2} + \pars{c/d}^{2}}}} \end{align}

Answer 3

If you know some inner product properties, you can solve those problems very easily. Just notice that $ax + by + cz$ is the inner product of two vectors $A = (a,b,c)$ and $X = (x,y,z)$.

First, consider the "normalized" $A$ vector: $A' = \sigma\frac{A}{\|A\|}$, where $\sigma = \frac{d}{|d|}$ is the sign of $d$. The equation $A \cdot X = d$ becomes $A' \cdot X = d'$, where $d' = \frac{d}{\sigma\|A\|}$. That is, $d' = \frac{|d|}{\|A\|}$. In this case, how do you read $$ P = \{X \,|\; A' \cdot X = d'\}? $$ $P$ is the set of points such that its projection in the $A'$ direction is always $d'$. Some people find it obvious that this is a plane.

Take any point $K \in P$. Now, $$ P = \{K+X \,|\; A' \cdot X = 0\}. $$ This is a plane passing through $K$, normal to $A'$! You take one point $K$, and add to it any vector that is perpendicular to $A'$. Some people like to think of a plane in this fashion.

In particular, if you consider the point $K = d'A'$, then $K \in P$. And now, $$ P = \{K+X \,|\; K \cdot X = 0\}. $$ That is, every point in $P$ is of the form $K + X$, where $K$ and $X$ are orthogonal.

(a) What is the distance of $P$ to the origin?
(b) What vector in $P$ is closest to the origin? That is, which vector in $P$ has smallest norm?

If you agree that it is the smallest norm amongst the points in $P$, then you agree that it is the smallest value of $$ \|K+X\| = \sqrt{\|K\|^2 + \|X\|^2} \leq \|K\|. $$ That is, $\|K\| = d'$ is the distance from $P$ to the origin. Since $K$ is orthogonal to the plane and $\|K\| = d'$.

(c) Consider the plane given by the equation in the primed form of part (b). Why is each plane containing the origin described by two distinct equations of this form? Why does each plane that does not contain the origin have a unique equation of this form?

The primed form $A' \cdot X = d'$ is given by a $A'$, a unitary vector normal to $P$ and $d' \geq 0$.

If you multiply by $-1$, and notice that $d' \geq 0$, you will get $(-A') \cdot X = -d'$. But this has the primed form only if $-d \geq 0$. That is, only if $d' = 0$. Since the only unitary vectors normal to $P$ are $A'$ and $-A'$, you get item (c).

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In your answer to (b), you don't need to expand $d'$. But it would be nice if you notice that $d'\vec{\mathbf{n}}$ is a point in $P$.