ex81 said:
I would assume that the x, y, and z are vectors.
As to the dot product, I have no clue. It would have been nice if it was ever covered...
One way to evaluate the dot product of two vectors
A and
B (boldface denotes vectors) is to evaluate it component-wise. In other words, if:
A = A
xi + A
yj + A
zk
B = B
xi + B
yj + B
zk
where i, j, and k are
unit vectors (vectors of magnitude 1) in the x, y, and z directions respectively, and the subscripted quantities are x, y, and z components, then the dot product is:
A · B = (A
xi + A
yj + A
zk)
· (B
xi + B
yj + B
zk)
Now, just using the ordinary distributive property of multiplication to expand out this product of two trinomials, you'd get
nine terms in total. However, most of these terms (in fact all of the so called "cross" terms) vanish. The reason is that (once again, due to the definition of the dot product) the dot product of two perpendicular vectors is 0. Hence
i · j = 0,
j · k = 0,
i · k = 0, and all permutations thereof. This means only three terms in the product survive, namely the three terms in which
like components are multiplied together (x with x, y with y, and z with z):
A · B = A
xB
xi·i + A
yB
yj·j + A
zB
zk·k
Now, another consequence of the definition of a dot product is that the dot product of a vector with itself is just equal to the magnitude of that vector, squared. Hence,
i·i = |
i|
2 etc. But since these are unit vectors, their magnitudes are 1, and we have
i·i =
j·j =
k·k = 1. Therefore, the result of the dot product is:
A · B = AxBx + AyBy + AzBz
THIS equation above is the one you should remember about how to compute the dot product of two vectors given their components. Everything above it was just an explanation of where it comes from. Notice that the result of the dot product of two vectors is a scalar. That's why it's also known as the scalar product. It's the type of vector multiplication that results in a scalar, as opposed to that other type of vector multiplication called the "cross product" or "vector product", which results in a vector. Of course, all of this is math you should already have been taught before being assigned physics problems like this one.