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Suppose P is a point in the plane ax + by + cz = d. Then the least distance from any point Q to...

Question:

Suppose {eq}\, P \, {/eq} is a point in the plane {eq}\, ax \, + \, by \, + \, cz \, = \, d {/eq}. Then the least distance from any point {eq}\, Q \, {/eq} to the plane equals the length of the orthogonal projection of {eq}\, \overrightarrow{PQ} \, {/eq} onto the normal vector {eq}\, \mathbf{n} \, = \, \left \langle a, \, b, \, c \right \rangle {/eq}.

a. Use this information to show that the least distance from {eq}\, Q \, {/eq} to the plane is {eq}\, \dfrac{\left | \overrightarrow{PQ} \, \cdot \, \mathbf{n} \right |}{\left | \mathbf{n} \right |} {/eq}.

Distance From a Point to a Plane

In vector geometry there is a well known formula for the distance {eq}D {/eq} from a point {eq}\textbf{Q}=(x_1,y_1,z_1) {/eq} to a plane whose equation is {eq}ax+by+cz=d {/eq} where {eq}a,b,c {/eq} and {eq}d {/eq} are constants. The distance formula is

{eq}\begin{eqnarray*} D=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}. \end{eqnarray*} {/eq}

The vector {eq}\textbf{n}=(a,b,c) {/eq} is a vector that is normal to the plane. Any vector {eq}\textbf{v}=c~\textbf{n} {/eq} where {eq}c\neq0 {/eq} is a scalar, that is, {eq}\textbf{v} {/eq} is a nonzero scalar multiple of {eq}\textbf{n} {/eq}, is also a normal vector for the plane. Note that the distance formula can also be written as {eq}D=\frac{|\textbf{n}\cdot\textbf{Q}-d|}{\|\textbf{n}\|} {/eq} where {eq}\cdot {/eq} is the dot product. When stating a point {eq}\textbf{Q} {/eq} is at a distance {eq}D {/eq} from a plane, it is understood that {eq}D {/eq} is the minimal distance the point {eq}\textbf{Q} {/eq} is from the plane.

Answer and Explanation:

The question is restated with slightly different notation. Suppose {eq}\textbf{P}=(x_0,y_0,z_0) {/eq} is a point on the plane {eq}ax+by+cz=d {/eq}. Then the least distance from any point {eq}\textbf{Q}=(x_1,y_1.z_1) {/eq} to the plane equals the length of the orthogonal projection of {eq}\textbf{u}=\overrightarrow{\textbf{PQ}} {/eq} onto the normal vector {eq}\textbf{n}=(a,b,c) {/eq}. Use this information to show that the least distance {eq}D {/eq} from {eq}\textbf{Q} {/eq} to the plane is

{eq}\begin{eqnarray*} D &=& \frac{|\textbf{u}\cdot\textbf{n}|}{\|\textbf{n}\|} \\ &=& \frac{|\overrightarrow{\textbf{PQ}}\cdot\textbf{n}|}{\|\textbf{n}\|}. \end{eqnarray*} {/eq}

In vector geometry there is a well known formula for the distance {eq}D {/eq} from a point {eq}\textbf{Q}=(x_1,y_1,z_1) {/eq} to a plane whose equation is {eq}ax+by+cz=d {/eq} where {eq}a {/eq}, {eq}b {/eq}, {eq}c {/eq} and {eq}d {/eq} are constants. The formula is given by the following equation

{eq}\begin{eqnarray*} D=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}. \end{eqnarray*} {/eq}

It is understood that the distance from a point to a plane is the minimal distance from the point to the plane.

Consider the vector projection of the vector {eq}\textbf{u}=\overrightarrow{\textbf{PQ}} {/eq} onto the normal vector {eq}\textbf{n}=(a,b,c) {/eq}. By definition, this is the vector

{eq}\begin{eqnarray*} \textbf{v} &=& \left(\frac{\textbf{u}\cdot\textbf{n}}{\|\textbf{n}\|^2}\right)~\textbf{n} \\ &=& \left(\frac{\overrightarrow{\textbf{PQ}}\cdot\textbf{n}}{\|\textbf{n}\|^2}\right)~\textbf{n}. \end{eqnarray*} {/eq}

Consequently

{eq}\begin{eqnarray*} \|\textbf{v}\| &=& \left|\frac{\textbf{u}\cdot\textbf{n}}{\|\textbf{n}\|^2}\right|~\|\textbf{n}\| \\ &=& \left|\frac{\overrightarrow{\textbf{PQ}}\cdot\textbf{n}}{\|\textbf{n}\|^2}\right|~\|\textbf{n}\| \\ &=& \frac{|\overrightarrow{\textbf{PQ}}\cdot\textbf{n}|}{\|\textbf{n}\|^2}~\|\textbf{n}\| \\ &=& \frac{|\overrightarrow{\textbf{PQ}}\cdot\textbf{n}|}{\|\textbf{n}\|} \\ &=& \frac{|\overrightarrow{\textbf{PQ}}\cdot\textbf{n}|}{\|(a,b,c)\|} \\ &=& \frac{|\overrightarrow{\textbf{PQ}}\cdot\textbf{n}|}{\sqrt{a^2+b^2+c^2}}. \end{eqnarray*} {/eq}

Since the point {eq}\textbf{P}=(x_0,y_0,z_0) {/eq} is a point on the plane {eq}ax+by+cz=d {/eq} then {eq}ax_0+by_0+cz_0=d {/eq}. Consider

{eq}\begin{eqnarray*} \overrightarrow{\textbf{PQ}}\cdot\textbf{n} &=& (\textbf{Q}-\textbf{P})\cdot\textbf{n} \\ &=& ((x_1,y_1,z_1)-(x_0,y_0,z_0))\cdot\textbf{n} \\ &=& (x_1,y_1,z_1)\cdot\textbf{n}-(x_0,y_0,z_0)\cdot\textbf{n} \\ &=& (x_1,y_1,z_1)\cdot(a,b,c)-(x_0,y_0,z_0)\cdot(a,b,c) \\ &=& ax_1+by_1+cz_1-(ax_0+by_0+cz_0) \\ &=& ax_1+by_1+cz_1-d. \end{eqnarray*} {/eq}

Consequently

{eq}\begin{eqnarray*} |\overrightarrow{\textbf{PQ}}\cdot\textbf{n}| &=& |ax_1+by_1+cz_1-d|. \end{eqnarray*} {/eq}

Therefore

{eq}\begin{eqnarray*} \|\textbf{v}\| &=& \frac{|\overrightarrow{\textbf{PQ}}\cdot\textbf{n}|}{\sqrt{a^2+b^2+c^2}} \\ &=& \frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}. \end{eqnarray*} {/eq}

This indicates that the length of the orthogonal projection of {eq}\textbf{u}=\overrightarrow{\textbf{PQ}} {/eq} onto the normal vector {eq}\textbf{n}=(a,b,c) {/eq} is equal to the value given by the distance formula from a point {eq}\textbf{Q} {/eq} to a plane, which is the minimal distance from the point {eq}\textbf{Q} {/eq} to the plane.


Learn more about this topic:

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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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