Distance of x,y and z intercepts of a plane

1. Apr 6, 2014

summer27

1. The problem statement, all variables and given/known data
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

2. Relevant equations

distance from a point to a plane: [tex]
d=|Ax0 + By0 + Cz0 +D|/√a2+b2+c2

3. 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)

Last edited: Apr 6, 2014
2. Apr 6, 2014

LCKurtz

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.)

3. Apr 6, 2014

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 $z= \frac{a^2b^2c}{b^2c^2+ a^2c^2+ a^2b^2}$.

Last edited by a moderator: Apr 6, 2014
4. Apr 6, 2014

Staff: Mentor

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