Geometry Please help !?

Consider a cube ABCDEFGH where J,K and L are the respective centers of the faces (ABEF), (EFGH) and (BCGF). F is equidistan to points J,K and L, D is equidistants of points J,K and L. (please refer to the figure)

Prove FD is orthogonal of the plan (JKL).

I can't use vectors to prove this...

The only thing I could think of is the right bissector since F and D are equidistant but the distant from F to the plane is not equal to the distant from D to the plane....

This is new for me to work with 3 dimensions

Any help will be appreciated !

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3 Answers

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  • Duke
    Lv 7
    5 years ago
    Favorite Answer

    I didn't understand if you are seeking for a proof using vectors or not. Here is an elementary solution - it is sufficient to prove that FD is orthogonal to 2 intersecting lines from the plane JKL.

    You know of course that in every square the ratio side : diagonal = 1 : √2.

    Draw outside the cube its vertical cross-section - the rectangle BDHF (notations as in your picture) and connect D with F, connect B with K. Since K is the midpoint of HF we have:

    |FB| : |BD| = 1 : √2, but |KF| : |FB| = (√2/2) : 1 = 1 : √2,

    so the right triangles DBF and BFK are similar, their corresponding sides are perpendicular, hence DF is perpendicular to KB. It remains to see that JL is perpendicular to DF also. Indeed JL || AC (look at triangle ACF) and AC is perpendicular to the plane BDHF - the latter means that AC is perpendicular to every line, lying on this plane, in particular to DF. Hence DF ┴ BK and DF ┴ JL, i.e. DF ┴ plane(JKL), what completes the proof.

    Remark: the intersection point P in xyzzy's answer DOES NOT bisect DF, if you connect H with the midpoint of BD, you can easily prove, that

    |DP| : |PF| = 2 : 1.

    The midpoint of BD is on the following hexagonal cross-section:

    http://gauss.math.nthu.edu.tw/d2/dg06exe/9521502/%...

    P.S. Here is another proof, using vectors. Consider a coordinate system with origin D, x-axis along DA, y-axis along DC, z-axis along DH; let the edge length is 1,

    so A(1, 0, 0), C(0, 1, 0), H(0, 0, 1) and

    [DF→] = [DA→] + [DC→] + [DH→], or [DF→](1,1,1)

    The equation of the plane JKL through B(1,1,0), E(1,0,1) and G(0,1,1) is:

    | x y z 1 |

    | 1 1 0 1 | = 0 (follow the link link below)

    | 1 0 1 1 |

    | 0 1 1 1 |

    http://en.wikipedia.org/wiki/Plane_%28geometry%29#...

    The determinant leads to x + y + z - 2 = 0, but if a plane has an equation

    ax + by + cz + d = 0, then (a, b, c) are coordinates of a vector, normal to the plane (see article), hence [DF→](1, 1, 1) ┴ plane JKL { x + y + z - 2 = 0 }

  • xyzzy
    Lv 7
    5 years ago

    How I would approach this..

    Identify a point point P where FD intersects the plane containing JKL.

    Now you need to prove that triangles JPF and KPF are right triangles.

    DJF is an isosceles triangle.

    P bisects DF

    JPF is a right triangle

    and repeat for KPF

    • Fred
      Lv 7
      5 years agoReport

      You started out fine, up to proving those two right triangles. If that can be shown, with both rt angles at P, then you have FD perp. to two intersecting lines in the plane, which makes FD perp. to the plane. But your next 2 statements are incorrect: DJF is not isosceles; P does not bisect DF.

  • 5 years ago

    There is a triad axis through FD.

    What is a triad axis? It is a line through a figure such that if you rotate the figure around that line by 120º then it will be unaltered.

    Proving that is very easy.

    HMmmmm ......

    I thought that it would be very easy to use this fact to prove that FD is orthogonal of the plane ... if I mange it I will edit this :)

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