Why am I getting t is a vector when solving for t in the vector equation of a 3d line. So as we all know. The vector equation of a 3d line is:
$$\vec{r} = \vec{r_0} + t\vec{v}$$
$$\left<x, y, z \right> = \left<x_0, y_0, z_0 \right> + t\left<a, b, c \right>$$
Now, when broken down...
$$\left<x-x_0, y-y_0, z-z_0 \right> = t\left<a, b, c \right>$$
$$\frac{\left<x-x_0, y-y_0, z-z_0 \right>}{\left<a, b, c \right>} = t$$
$$t = \frac{\left<x-x_0, y-y_0, z-z_0 \right>}{\left<a, b, c \right>}$$
I get $t$ is equal to a vector and I don't know why.
$$ t = \left<\frac{x-x_0}{a}, \frac{y-y_0}{b}, \frac{z-z_0}{c} \right> $$
Doesn't $t$ have to be a scalar? Why am I getting a vector when solving for it? I don't understand how we could write the symmetric equation of a 3D line like the following if $t$ were equal to a vector and not a scalar.
$$ t = \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$$

The trouble comes when you go from
$$(x-x_0, y-y_0, z-z_0) = t(a,b,c)$$
to
$$t = \frac{(x-x_0,y-y_0,z-z_0)}{(a,b,c)} = \big(\frac{x-x_0}{a}, \frac{y-y_0}{b}, \frac{z-z_0}{c}\big).$$
We cannot divide vectors in this way, and in fact there is no notion of vector division at all.
To see how you get the final expression for $t$, note that we have the equation
$$\left[\begin{array}{c} x \\ y \\ z \end{array} \right] = \left[\begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right] + t \left[\begin{array}{c} a \\ b \\ c \end{array} \right],$$
which is really the three equations
$$x = x_0+at$$
$$y=y_0+bt, \text{ and }$$
$$z=z_0+ct.$$
Now solve for $t$ in each case.
