Using the cross product to find the angle between two vectors in $\Bbb R^3$ 
Let $$u = \langle 1, −2, 3 \rangle \qquad \text{and} \qquad v = \langle −4, 5, 6 \rangle.$$
Find the angle between $u$ and $v$, first by using the dot product and then using the cross product.

I used the formula:
$U \cdot V = ||u|| \, ||v|| \cos \Delta$
and got $83^\circ$ from the dot product.
However, I am lost as how to use the cross product to find the answer.
 A: Hint The cross product satisfies
$$||{\bf a} \times {\bf b}|| = ||{\bf a}|| \, ||{\bf b}|| \sin \theta,$$
where $\theta \in [0, \pi]$ is the angle between $\bf a$ and $\bf b$.
(In fact this property is very nearly one of the common definitions of the cross product; see, e.g.,  Defn. 7.4 of Dennis G. Zill, Michael R. Cullen (2006). Advanced engineering mathematics (3rd ed.). Jones & Bartlett Learning.)
A: Let the angle between $u$ & $v$ be $\alpha$, now the dot product is given as follows $$u\cdot v=|u||v|\cos \alpha$$ $$\implies \cos \alpha=\frac{u\cdot v}{|u||v|}$$ $$=\frac{(1, -2, 3)\cdot (-4, 5, 6)}{|(1, -2, 3)||(-4, 5, 6)|}$$ $$\cos \alpha=\frac{-4-10+18}{\sqrt{(1)^2+(-2)^2+(3)^2}\sqrt{(-4)^2+(5)^2+(6)^2}}$$ $$\cos \alpha=\frac{4}{\sqrt{14\times 77}}$$$$\implies \color{blue}{\alpha=\cos^{-1}\left(\frac{4}{7\sqrt{22}}\right)\approx 83^\circ}$$
Cross product $|u\times v|$ is given as follows $$|u\times v|=\left|\begin{matrix} i&j&k\\ 1&-2&3\\-4&5&6
\end{matrix}\right|$$ $$=|-27i-18j-3k|$$ $$=\sqrt{(-27)^2+(-18)^2+(-3)^2}=\sqrt{1062}=3\sqrt{118}$$ Now, using cross product, the angle $\alpha$ between $u$ & $v$ is given as follows $$u\times v=|u||v|\sin \alpha(\hat n)$$ $$\implies |u\times v|=||u||v||\sin \alpha$$
$$\implies \sin\alpha=\frac{|u\times v|}{|u||v|}$$ $$=\frac{3\sqrt{118}}{7\sqrt{22}}=\frac{3}{7}\sqrt{\frac{59}{11}}$$
$$\color{blue}{\alpha=\sin^{-1}\left(\frac{3}{7}\sqrt{\frac{59}{11}}\right)\approx 83^\circ}$$
Your answer is correct
