Module 3: Now You Try II

1.  Compute the distance between the point $(1,2,3)$ and the line

$\begin{pmatrix}2\\2\\0\\\end{pmatrix} + t \begin{pmatrix}1\\-3\\5\\\end{pmatrix}.$

Solution:

The point $\vec{q}$ in the line to find is $\begin{pmatrix}12/5\\4/5\\2\\\end{pmatrix}.$ So the distance will be $\left\| \begin{pmatrix}1\\2\\3\\\end{pmatrix}- \begin{pmatrix}12/5\\4/5\\2\\\end{pmatrix} \right\| = \sqrt{\frac{22}{5}}.$

2.  Find the distance between the point $(2,8,5)$ and the plane given by $x-2y-2z=1$. 

Solution:

The line $\begin{pmatrix}2\\8\\5\\\end{pmatrix} + t \begin{pmatrix}1\\-2\\-2\\\end{pmatrix}$ will intersect the plane when $t$ is $25/9$. We use this to find the point of intersection between the line and the plane. We then compute the distance between this point and $\begin{pmatrix}2\\8\\5\\\end{pmatrix}$ , which is  $\frac{75}{9}$.