Day 52 in MAT229

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Today:

Announcements:
- Return of your homework (whoops! sorry, not yet):
  - 2
  - 22
  - 26
- Quiz Opportunity tomorrow will be over section 12.5
- Homework 12.3 is due tomorrow -- that's the last section covered for our last midterm exam on Thursday: it will cover everything up through 12.3 since our last exam.
Section 12.4: The Cross Product
- The next product we'll find convenient to define is the cross product, which is a product of two vectors resulting in a third vector, which juts into the space perpendicular to which the two vectors live.
  It is only defined (and useful) in three-space, which makes it somewhat unusual (the dot product exists and is useful in any dimensional space).
  The cross product is linear in its components: that is,
  
  ${(\bf{v}_1+\bf{v}_2)\times(\bf{w}_1+\bf{w}_2)}=\bf{v}_1\times\bf{w}_1+\bf{v}_1\times\bf{w}_2+\bf{v}_2\times\bf{w}_1+\bf{v}_2\times\bf{w}_2$
  This means that we can define it on the unit vectors in three space, and then deduce it using the component-wise definition of a three-vector.
  
  ${\hat{i}\times\hat{j}=\hat{k}}$
  ${\hat{j}\times\hat{k}=\hat{i}}$
  ${\hat{k}\times\hat{i}=\hat{j}}$
  
  When you reverse direction, however, you get the negative of the clockwise results:
  
  ${\hat{i}\times\hat{k}=-\hat{j}}$
  ${\hat{j}\times\hat{i}=-\hat{k}}$
  ${\hat{k}\times\hat{j}=-\hat{i}}$
  
  Furthermore, if you cross a vector with itself you get the zero vector:
  
  ${\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=\hat{k}\times\hat{k}=\bf{0}}$ Example:
  - #5, p. 838
  Now: here are the important geometric (rather than simply algebraic) properties of this product:
  
  Notice the difference in length between the dot- and cross-products:
  for the dot-product, it's the cosine; for the cross-product, it's the sine.
  
  There's another way of calculating the cross-product, which helps organize the calculation, and you may find it useful:
  
  So let me explain what's going on....
  The "absolute values" in this case represent determinants, which, for a 2x2 system with first row a, b and second row c, d is given by ${ad-bc}$ . So the computation above works out to
  
  ${\bf{v}\times\bf{w}=\bf{i}(b_1c_2-c_1b_2)-\bf{j}(a_1c_2-c_1a_2)+\bf{k}(a_1b_2-b_1a_2)}$
  Example:
  - #7, p. 838
  Examples:
  - #15, p. 838
  - #33, p. 839
  - Torque: #39
  - Torque: #41
  - A cross-product (and dot product) worksheet
Section 12.5: Equations of Lines and Planes
- If we have time, we'll approach this via a worksheet on lines and planes.
Links
- Some nice calculus worksheets

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