How to show that the line $\underset{\sim}{r}=-\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}+\mu\underset{\sim}{j}$ is tangent to a sphere?

Question

I want to show that $\underset{\sim}{r}=-\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}+\mu\underset{\sim}{j}$ is tangent to a sphere with equation $\left|\;\underset{\sim}{r}-\begin{bmatrix} -1 \\0\\2\\\end{bmatrix}\;\right|=3$.

So to start I converted the line into a more useful form, that is $\underset{\sim}{r}=\begin{bmatrix} -1 \\1\\-1\\\end{bmatrix}+\mu\begin{bmatrix} 0 \\1\\0\\\end{bmatrix}$.

Then to show tangency, I'm using the fact that tangents to a sphere are perpendicular to the line drawn from the centre to that point of contact. That is to say, for a line $\underset{\sim}{r}=\underset{\sim}{v}+\mu\underset{\sim}{m}$ that is tangent to the sphere $\left|\;\underset{\sim}{v}-\underset{\sim}{c}\;\right|=k$ the relationship $\underset{\sim}{m} \bullet (\underset{\sim}{v}-\underset{\sim}{c})=0$ holds (because the dot product of perpendicular lines is zero).

However when I do $\underset{\sim}{m} \bullet (\underset{\sim}{v}-\underset{\sim}{c})$, which in my case is $\begin{bmatrix} 0 \\1\\0\\\end{bmatrix}\bullet \left(\begin{bmatrix} -1 \\1\\-1\\\end{bmatrix}-\begin{bmatrix} -1 \\0\\2\\\end{bmatrix} \right)$, I get $1$ instead of $0$.

So what is wrong with my working (or maybe is there a problem with the question)?

Answer 1

Let the point vector $\underset{\sim}{r_0}=\begin{bmatrix} -1 \\1\\-1\\\end{bmatrix}+\mu\begin{bmatrix} 0 \\1\\0\\\end{bmatrix}$ lie on the sphere. Then,

$\left|\;\underset{\sim}{r_0}-\begin{bmatrix} -1 \\0\\2\\\end{bmatrix}\;\right|=3$

$\implies \mu=-1$.

Since, there is only value of $\mu$ satisfying (or a same-valued double root), the line touches the sphere. Hence it is a tangent at the point $\underset{\sim}{r_0} = (-1,0,-1)$.