Lines in 3D

Try to complete the following questions for Monday’s lesson.

1. Find the distance of the point  \(A(-1,1,2)\) to the line \[\displaystyle{\vec{r}=\begin{bmatrix}1\\-2\\3\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}}\]
2. The lines below intersect at a point \(A\). Use an algebraic method to find the coordinates of  \(A\), then verify your answer using GeoGebra.
   \[L_1: \vec{r}=\begin{bmatrix}-1\\-1\\4\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}\] \[L_2: \vec{r}=\begin{bmatrix}-2\\0\\7\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}\]
3. In 2D, any two non-parallel lines will have a point of intersection. In 3D, two lines can be non-parallel and have no point of intersection; such lines are called skew lines.
   Show that \(L_1\) from question 2 and \(L_3\) below are skew lines.
   \[L_3: \vec{r}=\begin{bmatrix}-2\\5\\12\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}\]