You'll usually do dot product calculations with the vectors in component form. Let's look first at some simple dot products of the vectors i, j and k with each other.

In 2-space, since i = [1, 0] and j = [0, 1], we get

i•i = 1,   j•j = 1   and   i•j = 0

In 3-space, since i = [1, 0, 0], j = [0, 1, 0] and k = [0, 0, 1], we get

i•i = j•j = k•k = 1   and   i•j = j•k = k•i= 0.

In words, the dot product of i, j or k with itself is always 1, and the dot products of i, j and k with each other are always 0.

The dot product of a vector with itself is a sum of squares:

• in 2-space, if u = [u1, u2] then u•u = u12 + u22,
• in 3-space, if u = [u1, u2, u3] then u•u = u12 + u22 + u32.

This means that u•u is generally positive; the only possible exception occurs when all of the components of u are 0, i.e. when u = 0. Thus:

0•0 = 0, and if u ≠ 0, u•u > 0.

Notice also that the sum of squares in each case is the same sum of squares that appears in the formulas for the lengths of vectors in 2-space and 3-space (without the square root sign). This gives you a relation between dot products and the length of a vector:

u•u = ||u||2.

Dot products are commutative: for vectors u and v (both either in 2-space or in 3-space), u•v = v•u.

Dot products are distributive over addition: for vectors u, v and w (all either in 2-space or in 3-space), u•(v + v) = u•v + u•w.

Both of these rules are easy to check (use the component form of the definition of the dot product) .

When finding the dot product of scalar multiples of two vectors, you can multiply by the scalars either before or after you find the dot product: for any scalar c and vectors u and v,

(cu)•v = u•(cv) = c(u•v).

This is again easy to check using components.

Here's a list summarizing the calculation rules for dot products.

For any vectors u, v and w all in 2-space or all in 3-space and any scalar c,

• i•i = j•j = k•k = 1   and   i•j = j•k = k•i= 0.
• 0•0 = 0, and if u ≠ 0, u•u > 0
• u•u = ||u||2
• u•v = v•u
• u•(v + w) = u•w + v•w
• (cu)•v = u•(cv) = c(u•v).