Questions of calculus of crystal structures

mysci
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upload_2015-3-3_22-26-6.png


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.
 
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mysci said:
View attachment 79881

We know the rule of cross productView attachment 79882 or
2d308f37dd82911690b919157eace04d.png


Why here |absinΘ| = View attachment 79883 , and View attachment 79884 = c cos Φ in the above picture?
|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.
 
The base is formed by two vectors with lengths ||a|| and ||b|| and angle between them \theta. 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 height/||b||= sin(\theta) and height= ||b|| sin(\theta). The area of a parallelogram is "height times base" so ||a||||b|| sin(\theta).
 
Mark44 said:
|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.

Thanks.

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.

Thanks.
 
mysci said:
Thanks.

Yes, but why not |axb|, is |axb|(unit vector n) in third step?
|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##.
mysci said:
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.

Thanks.
 
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.
 
Mark44 said:
|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##.

By the way, how do you type the vector symbol in here? I can't find this symbol. Thanks.
 
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}##
 
Mark44 said:
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}##
Thanks.

Then
I got following,
absinΘ = |axb|
ccosΦ = n·c
Is it right?
 
  • #10
mysci said:
Thanks.

Then
I got following,
absinΘ = |axb|
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).
mysci said:
ccosΦ = n·c
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.
mysci said:
Is it right?
 
  • #11
Mark44 said:
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.
Thank you.:wink:
 
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