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Trigonometry Question

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A drawing shows sodium and chloride ions positioned at the corners of a cube that is part of the crystal structure of sodium chloride. The edge of the cube is 0.281nm in length. Find the distance between the sodium ion located at one corner of the cube and the chloride ion located on the diagonal at the opposite corner.

    2. Relevant equations
    I'm thinking pythagorean theorem and one or more of the trig functions perhaps?

    3. The attempt at a solution
    Well...I figured the diameter if the cube is drawn inside a circle is 2(0.281)=0.562nm.
    Not sure what to do this...it's difficult for me to conceptualize.

    (The answer in the book is 0.487nm)
  2. jcsd
  3. Oct 9, 2009 #2


    Staff: Mentor

    One way to approach this is as a distance problem in 3-D space. Put on corner of the cube at the origin - (0, 0, 0) - and the opposite corner at (.281, .281, .281).

    The distance d from a point (x1, y1, z1) to another point (x2, y2, z2) is
    [tex]d~=~\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}[/tex]

    BTW, I get a distance of 0.48670627692685451948121242196315 nm.
  4. Oct 9, 2009 #3
    I kind of understand, but what would this look like as a picture on paper? Imagine no numbers...
  5. Oct 9, 2009 #4


    Staff: Mentor

    Unfortunately, I am unable to draw you a picture. The distance formula in my previous post is the 3-D counterpart to the length of the hypotenuse in the Theorem of Pythagoras.

    I'll try to describe what this formula does in words, and leave the drawing to you. In your NaCl cube, let's say that you want to find the distance from the lower left corner (O) of the front face, to the upper right corner (P) of the rear face. To find this distance, consider the triangle whose vertices are O, P, and the point at the lower right corner of the front face (call this point Q). Triangle OPQ is a right triangle, with the right angle formed by sides OQ and OP. By the theorem of Pythagoras, |OP| = sqrt(|OQ|2 + |QP|2). |OQ| is given in your problem, and |OQ| is the length of the diagonal across one of the square faces.

    Hopefully I have described this well enough so that you can draw the picture.
  6. Oct 9, 2009 #5
    Thank you for this very detailed help. I appreciate the effort you put into helping me construct a picture if you will. It does make sense to me after I drew it based on your instructions. Thanks again.
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