Module 3: Now You Try III

Let's find a plane which contains $\ell_1$. It will have normal
$\begin{pmatrix}1\\1\\1\\\end{pmatrix} \times \begin{pmatrix}1\\1/2\\1/3\\\end{pmatrix} = \begin{pmatrix}-1/6\\2/3\\-1/2\\\end{pmatrix}$ Since $\ell_1$ passes through the origin our plane must be:
$\frac{-1}{6} x + \frac{2}{3}y - \frac{1}{2}z = 0.$Instead we will make the plane $x-4y+3z=0$ (Why is this ok?). Line $\ell_2$ has the point $\begin{pmatrix}1\\0\\0\\\end{pmatrix}$ in it so the line through this point perpendicular to the plane is $\begin{pmatrix}1\\0\\0\\\end{pmatrix} + t \begin{pmatrix}1\\-4\\3\\\end{pmatrix}.$

Solve for the intersection point to find that $t = -1/26$ as follows:
$1+t -4(-4t)+3(3t) = 1+26t = 0.$
Compute the distance between the intersection point and $\begin{pmatrix}1\\0\\0\\\end{pmatrix}$ to get $\sqrt{26}$.