# Questions of calculus of crystal structures

1. Mar 3, 2015

### mysci

We know the rule of cross product or

Why here |absinΘ| = , and = c cos Φ in the above picture?

Thanks for explanation.

2. Mar 3, 2015

### 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.

3. Mar 3, 2015

### HallsofIvy

Staff Emeritus
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 $||a||||b|| sin(\theta)$.

4. Mar 3, 2015

### mysci

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.

5. Mar 3, 2015

### 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 $\phi$.

6. Mar 3, 2015

### mysci

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.

7. Mar 4, 2015

### mysci

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

8. Mar 4, 2015

### 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 $\vec{a}$

9. Mar 4, 2015

### mysci

Thanks.

Then
I got following,
absinΘ = |axb|
ccosΦ = n·c
Is it right?

10. Mar 4, 2015

### 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.

11. Mar 4, 2015

Thank you.