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Distance of x,y and z intercepts of a plane

  1. Apr 6, 2014 #1
    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. jcsd
  3. Apr 6, 2014 #2

    LCKurtz

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    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.)
     
  4. Apr 6, 2014 #3

    HallsofIvy

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    So the equation of the plane is [itex]f(x, y, z)= \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= 1[/itex]. The distance from the origin to the plane will be minimum when the vector perpendicular to the plane, [itex]\nabla f(x,y,z)= \frac{1}{a}\vec{i}+ \frac{1}{b}\vec{j}+ \frac{1}{b}\vec{k}[/itex], is parallel to the vector from the origin to the point, [itex]x\vec{i}+ y\vec{j}+ z\vec{k}[/itex]. That is, there must be some number, [itex]\lambda[/itex] such that [itex]x= \frac{\lambda}{a}[/itex]
    [itex]y= \frac{\lambda}{b}[/itex], and [itex]z= \frac{\lambda}{c}[/itex]. (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 [itex]\frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{\lambda}{a^2}+ \frac{\lambda}{b^2}+ \frac{\lambda}{c^2}= 1[/itex]. That is, [itex]\lambda\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)= 1[/itex] so [itex]\lambda= \frac{1}{\frac{1}{a^2}+ \frac{1}{b^2}+ \frac{1}{c^2}}[/itex].

    Of course, [itex]\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}[/itex] so that [itex]\lambda= \frac{a^2b^2c^2}{b^2c^2+ a^2c^2+ a^2b^2}[/itex]. Putting that into [itex]x= \frac{\lambda}{a}[/itex], [itex]y= \frac{\lambda}{b}[/itex], and [itex]z= \frac{\lambda}{c}[/itex] gives
    [itex]x= \frac{ab^2c^2}{b^2c^2+ a^2c^2+ a^2b^2}[/itex], [itex]y= \frac{a^2bc^2}{b^2c^2+ a^2c^2+ a^2b^2}[/itex], and [itex]z= \frac{a^2b^2c}{b^2c^2+ a^2c^2+ a^2b^2}[/itex].
     
    Last edited by a moderator: Apr 6, 2014
  5. Apr 6, 2014 #4

    Mark44

    Staff: Mentor

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