Equation 1
$$\vec{a} * \vec{b} = (|\vec{a}| | \vec{b}| sin θ)ň $$
Equation 2
$$\vec{a} * \vec{b} = (a_yb_z - a_zb_y)î - (a_xb_z - a_zb_x)ĵ + (a_xb_y - a_yb_x)k̂ $$
I have managed to figure out and understand how to get all the terms within the brackets. However, I do not understand how to get the + and - symbols between the brackets.
Additional Information
$$ î * ĵ = k̂ $$
$$ ĵ * k̂ = î $$
$$ k̂ * î = ĵ $$
$$ ĵ * î = -k̂ $$
$$ k̂ * ĵ = -î $$
$$ î * k̂ = -ĵ $$
$$ î * î = ĵ * ĵ = k̂ * k̂ = 0 $$
Thank you!
Further Update
So, to begin...
$$\vec{a} = a_xî+a_yĵ+a_zk̂$$
$$\vec{b} = b_xî+b_yĵ+b_zk̂$$
Multiplying out vector a and b gives 9 terms:
$$a_xîb_xî + a_xîb_yĵ + a_xîb_zk̂ + a_yĵb_xî + a_yĵb_yĵ + a_yĵb_zk̂ + a_zk̂b_xî + a_zk̂b_yĵ + a_zk̂b_zk̂ $$
The 3 terms in which the unit vectors (î, ĵ and k̂) multiply by themselves, are equal to zero, and hence disappear. For example, the following will disappear:
$$ a_xîb_xî $$
Therefore, we are left with 6 terms: $$a_xîb_yĵ + a_xîb_zk̂ + a_yĵb_xî + a_yĵb_zk̂ + a_zk̂b_xî + a_zk̂b_yĵ $$
Next, I use the information under additional information to get:
$$a_xb_yk̂ + a_xb_z(-ĵ) + a_yb_x(-k̂) + a_yb_zî + a_zb_xĵ + a_zb_y(-î) $$
Now, I group the like terms together by the unit vector.
$$ (a_xb_y - a_yb_x)k̂ + (a_xb_z-a_zb_x)ĵ + (a_yb_z - a_zb_y)î $$
The above line is not correct (but what I keep getting). The book states equation 2 is what I should get. Where have I gone wrong?
Thank you
Define $a=(a_1,a_2,a_3),b=(b_1,b_2,b_3).$
I'll assume you are familiar with the dot product identity (otherwise, prove it by the law of cosines):
$$a \cdot b=|a| |b| \cos \theta.$$
Note that from your equation 2,
$$|a \times b|^2=(a_2b_3-b_2a_3)^2+(b_1a_3-a_1b_3)^2+(a_1b_2-b_1a_2)^2.$$
Some clever reorganization of this messy expression will get you the following:
$$|a \times b|^2=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2,$$
which can now be written in terms of the dot product as
$$|a \times b|^2=|a|^2|b|^2-(a\cdot b)^2=|a|^2|b|^2-|a|^2|b|^2\cos^2 \theta=|a|^2|b|^2\sin^2\theta$$
Square root both sides and conclude.