Q: The lines $L1$ and $L2$ have equations
$$L1: \mathbf{r} = \begin{bmatrix}1\\-2\\1\end{bmatrix} + s\begin{bmatrix}2\\1\\-2\end{bmatrix}$$
and
$$L2: \mathbf{r}= \begin{bmatrix}-3\\-11\\5\end{bmatrix} + t\begin{bmatrix}-1\\3\\1\end{bmatrix}$$
Calculate the sine of the angle between $L1$ and $L2$:
My solution:
$$\mathbf{a}=\begin{bmatrix}2\\1\\-2\end{bmatrix}\quad\text{and}\quad \mathbf{b} = \begin{bmatrix}-1\\3\\1\end{bmatrix},$$
$$|\mathbf{a}|=\sqrt{2^2 + 1^2 + (-2)^2}=\sqrt{4 + 1 + 4}=\sqrt{9}=3,$$
$$|\mathbf{b}|=\sqrt{(-1)^2 + 3^2 + 1^2}=\sqrt{1 + 9 + 1}= \sqrt{11},$$
$$\mathbf{a}\times\mathbf{b}=\begin{bmatrix}2\\1\\-2\end{bmatrix}\times \begin{bmatrix}-1\\3\\1\end{bmatrix}=\begin{bmatrix}7\\0\\7\end{bmatrix}=\mathbf{n},$$
$$|\mathbf{n}|=\sqrt{(7^2 + 0^2 + 7^2}=\sqrt{49 + 0 + 49}=\sqrt{98}= 7\sqrt{2}.$$
Using $$\mathbf{a}\times\mathbf{b} =|\mathbf{a}|\,|\mathbf{b}|\sin\theta\,\hat{n}$$
then $$7\sqrt{2}=3\sqrt{11}\sin\theta$$
so $$\sin\theta=\frac{7\sqrt{2}}{3\sqrt{11}}=\frac{7\sqrt{22}}{33}.$$
Is My Solution Correct?
The Textbook Gives The Answer As $\sqrt{22}$