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Multivariable calculus equation of a plane

  1. Jul 17, 2013 #1
    equation of the form

    ax+bx+cx=a0+b0+c0

    ax+bx+cx=d


    what exactly does "d" represent? is it the distance of a point from the plane? or is it the shifting of the plane along the normal vector of the plane?
    and how does "d" represent that again?
     
  2. jcsd
  3. Jul 17, 2013 #2
    A way to write an equation for a plane in [itex]\mathbb{R}^3[/itex] is
    [tex]\vec{w} \cdot \vec{r} = d[/tex]and the plane contains the point [tex] d\frac{ \vec{w}}{\left\|\vec{w}\right\|^2}[/tex] You can check that the above is true. Now if [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] are parallel to the plane and one is not a scalar multiple of the other, then we could write the plane as [tex]\vec{r}(s,t) = d\frac{ \vec{w}}{\left\|\vec{w}\right\|^2} + s\vec{u} + t\vec{v}[/tex] since we can always parameterize the plane as a point on the plane plus scalar multiples of two spanning vectors added on.

    So you could think that the plane has been shifted by [itex] d\frac{ \vec{w}}{\left\|\vec{w}\right\|^2}[/itex], which is along the normal. It is somewhat ambiguous perhaps because some other shifts in other directions would result in the same plane.

    The above applies to planes in [itex]\mathbb{R}^3[/itex].
     
    Last edited: Jul 17, 2013
  4. Jul 17, 2013 #3

    LCKurtz

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    I guess you mean ##ax + by + cz = d##, not what you wrote. In that case the quantity$$\frac{|d|}{\sqrt{a^2+b^2+c^2}}$$gives the distance of the plane from the origin.
     
  5. Jul 17, 2013 #4

    HallsofIvy

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    Also, if x= y= 0 then cz= d so z= d/c, if x= z= 0 then by= d so y= d/b, and if y= z= 0 then ax= d so x= d/a. That is, (d/a, 0, 0), (0, d/b, 0), and (0, 0, d/c) are the x, y, and z intercepts of the plane.
     
  6. Jul 17, 2013 #5
    yes that's what i meant, could you show me how this yields the distance of the plane from the origin?
    what does d itself represent?
     
  7. Jul 17, 2013 #6

    LCKurtz

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    Look in your calculus book how to calculate the distance from a plane to a point and use it to calculate the distance to (0,0,0).

    ##d## by itself doesn't have any particular meaning. If you multiply the equation through by ##2## you will have the same plane but now ##2d## on the right side. So ##d## in relation to the coefficients is what is significant. And the formula above gives geometric meaning to it.
     
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