Let
\[\mathbf{a} = \begin{pmatrix} 5 \\ -3 \\ -4 \end{pmatrix} \quad \text{and} \quad \mathbf{b} = \begin{pmatrix} -11 \\ 1 \\ 28 \end{pmatrix}.\]There exist vectors $\mathbf{p}$ and $\mathbf{d}$ such that the line containing $\mathbf{a}$ and $\mathbf{b}$ can be expressed in the form
\[\mathbf{v} = \mathbf{p} + \mathbf{d} t.\]Furthermore, for a certain choice of $\mathbf{d}$, it is the case that for all points $\mathbf{v}$ lying on the same side of $\mathbf{a}$ that $\mathbf{b}$ lies on, the distance between $\mathbf{v}$ and $\mathbf{a}$ is $t$.  Find $\mathbf{d}$.
From the given property, the distance between $\bold{v}$ and $\bold{a}$ is 0 when $t = 0$, so $\bold{v} = \bold{a}$.  But the equation $\bold{v} = \bold{p} + \bold{d} t$ becomes
\[\bold{v} = \bold{p}\]when $t = 0$.  Hence, $\bold{p} = \bold{a}$, so the equation of the line is
\[\bold{v} = \bold{a} + \bold{d} t.\]Also, the vector $\bold{b}$ lies on the line, and the distance between $\bold{a}$ and $\bold{b}$ is
\[\|\bold{a} - \bold{b}\| = \left\| \begin{pmatrix} 5 \\ -3 \\ -4 \end{pmatrix} - \begin{pmatrix} -11 \\ 1 \\ 28 \end{pmatrix} \right\| = \left\| \begin{pmatrix} 16 \\ -4 \\ -32 \end{pmatrix} \right\| = \sqrt{16^2 + (-4)^2 + (-32)^2} = 36.\]Hence, the value of $t$ for which $\bold{b} = \bold{a} + \bold{d} t$ is $t = 36$, which means
\[\begin{pmatrix} -11 \\ 1 \\ 28 \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \\ -4 \end{pmatrix} + 36 \bold{d}.\]Isolating $\bold{d}$, we find
\[\bold{d} = \boxed{\begin{pmatrix} -4/9 \\ 1/9 \\ 8/9 \end{pmatrix}}.\]