Find the minimum distance from the point (0,0,0) to the plane (9x+4y+z=5).

Question:

Find the minimum distance from the point {eq}(0,0,0){/eq} to the plane {eq}(9x+4y+z=5){/eq}.

Shortest Distance Of Point From A Plane

For a given equation of plane {eq}Ax+By+Cz+D = 0 {/eq} , the shortest distance of the point {eq}(x_1,y_1,z_1) {/eq} from given plane is given by the following expression,

{eq}\begin{align*} d = \frac{|Ax_1+By_1+Cz_1 +D|}{\sqrt{A^2+B^2+C^2}}. \end{align*} {/eq}

Answer and Explanation: 1

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Given equation of plane is {eq}(9x+4y+z=5) {/eq} , comparing with {eq}Ax+By+Cz+D = 0 {/eq} , we get {eq}A=9,B=4,C=1,D=-5 {/eq} and

{eq}(x_1,y_1,z_...

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How to Find the Distance Between a Point & a Line

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Chapter 31 / Lesson 5
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Read about finding the distance from a point to a line. Learn about the distance from a point to a line formula and its application. Also, understand how to find the distance without a formula.


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