Distance of x,y and z intercepts of a plane

[summer27]

Homework Statement

If a,b and c are the x-,y-, and z-intercepts of a plane, respectively, and d is the distance from the origin to the plane show that:

1/d²=1/a²+1/b²+1/c²

Homework Equations

distance from a point to a plane: <br /> d=|Ax0 + By0 + Cz0 +D|/√a²+b²+c²<br /> <br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> So I have been able to draw it and what I am getting is that I can make 3 right angled triangles. One would be: (O is the origin, all of these would be vectors)<br /> AO +OD =DA<br /> OB+ BD = DO<br /> OC +CD =DO<br /> I am also thinking that the distance from the plane to the origin would have to be the normal of the plane. But that&#039;s all I have.<br /> (i&#039;m sorry i could not get the equation written with the code)

 

[LCKurtz]

Homework Statement

If a,b and c are the x-,y-, and z-intercepts of a plane, respectively, and d is the distance from the origin to the plane show that:

1/d²=1/a²+1/b²+1/c²

Homework Equations

distance from a point to a plane: <br /> d=\frac{|Ax_0 + By_0 + Cz_0 +D|}{\sqrt{A^2+B^2+C^2}}

The Attempt at a Solution

So I have been able to draw it and what I am getting is that I can make 3 right angled triangles. One would be: (O is the origin, all of these would be vectors)
AO +OD =DA
OB+ BD = DO
OC +CD =DO
I am also thinking that the distance from the plane to the origin would have to be the normal of the plane. But that's all I have.
(i'm sorry i could not get the equation written with the code)

Don't include sup tags and square root symbols in tex. Quote this to see how I fixed that fraction. Notice that the plane with those intercepts can be written as$$
\frac 1 a x + \frac 1 b y + \frac 1 c z = 1$$Now use your quoted formula to calculate the distance from $(0,0,0)$ to the plane. (I corrected your formula too.)

 

[HallsofIvy]

So the equation of the plane is f(x, y, z)= \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= 1. The distance from the origin to the plane will be minimum when the vector perpendicular to the plane, \nabla f(x,y,z)= \frac{1}{a}\vec{i}+ \frac{1}{b}\vec{j}+ \frac{1}{b}\vec{k}, is parallel to the vector from the origin to the point, x\vec{i}+ y\vec{j}+ z\vec{k}. That is, there must be some number, \lambda such that x= \frac{\lambda}{a}
y= \frac{\lambda}{b}, and z= \frac{\lambda}{c}. (This argument is equivalent to the "Lagrange multiplier method" for dealing with max-min problems.)

Of course, since the point, (x, y, z), is on the plane, it must also satisfy the equation of the plane so we must have \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{\lambda}{a^2}+ \frac{\lambda}{b^2}+ \frac{\lambda}{c^2}= 1. That is, \lambda\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)= 1 so \lambda= \frac{1}{\frac{1}{a^2}+ \frac{1}{b^2}+ \frac{1}{c^2}}.

Of course, \frac{1}{a^2}+\frac{1}{b^2}+ \frac{1}{c^2}= \frac{b^2c^2}{a^2b^2c^2}+ \frac{a^2c^2}{a^2b^2c^2}+ \frac{a^2b^2}{a^2b^2c^2}= \frac{b^2c^2+ a^2c^2+ a^2b^2}{a^2b^2c^2} so that \lambda= \frac{a^2b^2c^2}{b^2c^2+ a^2c^2+ a^2b^2}. Putting that into x= \frac{\lambda}{a}, y= \frac{\lambda}{b}, and z= \frac{\lambda}{c} gives
x= \frac{ab^2c^2}{b^2c^2+ a^2c^2+ a^2b^2}, y= \frac{a^2bc^2}{b^2c^2+ a^2c^2+ a^2b^2}, and z= \frac{a^2b^2c}{b^2c^2+ a^2c^2+ a^2b^2}.

 

[Mark44]

So the equation of the plane is f(x, y, z)= \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= 1. The distance from the origin to the plane will be minimum when the vector perpendicular to the plane, \nabla f(x,y,z)= \frac{1}{a}\vec{i}+ \frac{1}{b}\vec{j}+ \frac{1}{b}\vec{k}, is parallel to the vector from the origin to the point, x\vec{i}+ y\vec{j}+ z\vec{k}. That is, there must be some number, \lambda such that x= \frac{\lambda}{a}
y= \frac{\lambda}{b}, and z= \frac{\lambda}{c}. (This argument is equivalent to the "Lagrange multiplier method" for dealing with max-min problems.)

Of course, since the point, (x, y, z), is on the plane, it must also satisfy the equation of the plane so we must have \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{\lambda}{a^2}+ \frac{\lambda}{b^2}+ \frac{\lambda}{c^2}= 1. That is, \lambda\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)= 1 so \lambda= \frac{1}{\frac{1}{a^2}+ \frac{1}{b^2}+ \frac{1}{c^2}}.

Of course, \frac{1}{a^2}+\frac{1}{b^2}+ \frac{1}{c^2}= \frac{b^2c^2}{a^2b^2c^2}+ \frac{a^2c^2}{a^2b^2c^2}+ \frac{a^2b^2}{a^2b^2c^2}= \frac{b^2c^2+ a^2c^2+ a^2b^2}{a^2b^2c^2} so that \lambda= \frac{a^2b^2c^2}{b^2c^2+ a^2c^2+ a^2b^2}. Putting that into x= \frac{\lambda}{a}, y= \frac{\lambda}{b}, and z= \frac{\lambda}{c} gives
x= \frac{ab^2c^2}{b^2c^2+ a^2c^2+ a^2b^2}, y= \frac{a^2bc^2}{b^2c^2+ a^2c^2+ a^2b^2}, and z= \frac{a^2b^2c}{b^2c^2+ a^2c^2+ a^2b^2}.

This might be irrelevant to the OP, who after all, has posted in the Precalculus section.