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Questions of calculus of crystal structures

  1. Mar 3, 2015 #1

    We know the rule of cross product upload_2015-3-3_22-27-38.png or 2d308f37dd82911690b919157eace04d.png

    Why here |absinΘ| = upload_2015-3-3_22-39-31.png , and upload_2015-3-3_23-13-6.png = c cos Φ in the above picture?

    Thanks for explanation.
  2. jcsd
  3. Mar 3, 2015 #2


    Staff: Mentor

    |a||b|sinθ is the area of the base, and is also the magnitude of the cross product of ##\vec{a}## and ##\vec{b}## -- i.e., |##\vec{a}## X ##\vec{b}|##. |c|cosφ gives the height of the parallelipiped. The product of the area of the base and the height gives the volume of the cell.
  4. Mar 3, 2015 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    The base is formed by two vectors with lengths ||a|| and ||b|| and angle between them [itex]\theta[/itex]. Draw a line from the tip of line b to the base, the line forming the vector a. That gives you a right triangle with hypotenuse of length ||b||. So the "opposite side", the height of the parallelogram forming the base of the figure. The "opposite side over the hypotenuse" is sine of the angle so [itex]height/||b||= sin(\theta)[/itex] and [itex]height= ||b|| sin(\theta)[/itex]. The area of a parallelogram is "height times base" so [itex]||a||||b|| sin(\theta)[/itex].
  5. Mar 3, 2015 #4

    Yes, but why not |axb|, is |axb|(unit vector n) in third step?

    absinΘ and |axbl are also magnitudes, but |axb|(unit vector n) is a vector. absinΘ = |axb| ≠ |axb|(unit vector n) = vector a x vector b.

    However, here absinΘ = |axb|(unit vector n). I don't understand this.

    On the other hand, ccosφ is the height of parallelogram, how to change it to vector c? I don't understand it as well.

  6. Mar 3, 2015 #5


    Staff: Mentor

    |a x b| is a scalar, while |a x b|n is a vector that points straight up, and that whose magnitude is the area of the base. If you dot this vector (|a x b|n) with c, you get the volume. One definition for the dot product of a and b is ##a \cdot b = |a| |b| cos(\theta)##, where ##\theta## is the angle between the two vectors. In your problem, the angle is ##\phi##.
  7. Mar 3, 2015 #6
    I may get something.
    In fact,
    absinΘ = |axb|
    ccosΦ = n·c
    Is it right?

    I thought absinΘ = |axb|n and ccosΦ = c before I get the above thinking.
  8. Mar 4, 2015 #7
    By the way, how do you type the vector symbol in here? I can't find this symbol. Thanks.
  9. Mar 4, 2015 #8


    Staff: Mentor

    I use LaTeX. Put either two # symbols at the front and two more at the end (for inline) or two $ symbols front and back (for standalone).

    Here I'm adding an extra space between each pair so you can see what it looks like without being rendered: # #\vec{a}# #
    Removing the spaces gives ##\vec{a}##
  10. Mar 4, 2015 #9

    I got following,
    absinΘ = |axb|
    ccosΦ = n·c
    Is it right?
  11. Mar 4, 2015 #10


    Staff: Mentor

    Should be |a||b|sinθ = |a x b|. a and b are vectors, so ab is not defined. Both sides of the equation should be scalars, which is why you have the magnitudes (absolute values).
    The right side is a scalar because it's a dot product, so the left side needs to be a scalar as well.
    The left side should be |c|cosΦ.
  12. Mar 4, 2015 #11
    Thank you.:wink:
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