I understand that the cross product formula is
$$\mathbf{A} \times \mathbf{B} =\left\|\mathbf{A} \right\|\left\|\mathbf{B} \right\|\sin(\theta)\ \mathbf{n},$$
where $\mathbf{n}$ is a unit vector perpendicular to the plane containing $\mathbf{A}$ and $\mathbf{B}$ in the direction given by the right-hand rule.
I started doing the following exercise from chapter 1.1.2 Vector Algebra: Component Form of Introduction to Electrodynamics, fourth edition, by David J. Griffiths:
Example 1.2. Find the angle between the face diagonals of a cube
I begin by using a cube of side $1$, as shown here in the textbook:
Let $\mathbf{A} = (0, 1, 1)$ and $\mathbf{B} = (1, 0, 1)$.
Then the cross product formula gives us that
$$(1, 1, -1) = (\sqrt{2})(\sqrt{2}) \sin(\theta) \mathbf{n}.$$
So we need to solve for $\theta$, which means I need to deduce $\mathbf{n}$ first, right?
But then I remember: Isn't $\mathbf{A} \times \mathbf{B}$ itself supposed to be the "normal/perpendicular vector" to the plane in which $\mathbf{A}$ and $\mathbf{B}$ lie? So, in that case, how does the formula $\mathbf{A} \times \mathbf{B} =\left\|\mathbf{A} \right\|\left\|\mathbf{B} \right\|\sin(\theta)\ \mathbf{n}$ make sense? And, furthermore, how do we now find $\mathbf{n}$ in $(1, 1, -1) = (\sqrt{2})(\sqrt{2}) \sin(\theta) \mathbf{n}$? This is odd, because the cross product formula is actually often written as $\mathbf{A} \times \mathbf{B} =\left\|\mathbf{A} \right\|\left\|\mathbf{B} \right\|\sin(\theta)$ (with the absence of $\mathbf{n}$), no? As you can see, I became very confused.
I would greatly appreciate it if people would please take the time to clarify my confusion/thoughts here.
Best Answer
$\mathbf A=(0,1,1)$
$\mathbf B=(0,0,1)$
$\mathbf A\times\mathbf B=(1,1,-1)$
$\mathbf n=\dfrac{(1,1,-1)}{\|(1,1,-1)\|}=\left(\dfrac1{\sqrt3},\dfrac1{\sqrt3},-\dfrac1{\sqrt3}\right)$
Can you take it from here?