The position vectors of points \(A\), \(B\) and \(C\) are
\[\begin{equation*}
\mathbf{a} = 4\mathbf{i} - 9\mathbf{j} - \mathbf{k}, \quad \mathbf{b} = \mathbf{i} + 3\mathbf{j} + 5\mathbf{k}, \quad\text{and}\quad \mathbf{c} = p\mathbf{i} - \mathbf{j} + 3\mathbf{k}.
\end{equation*}\]
Find the unit vector parallel to the vector \(\mathbf{AB}\).
The vector \(\overrightarrow{AB}\) begins at \(A\) and ends at \(B\). We have the following diagram.
From the diagram, we have that \(\mathbf{a} + \overrightarrow{AB} = \mathbf{b}\), i.e. that
\[\begin{equation*}
\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) - (4\mathbf{i} - 9\mathbf{j} - \mathbf{k}) = -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}.
\end{equation*}\]
The length of this vector is
\[\begin{equation*}
\sqrt{(-3)^2 + 12^2 + 6^2} = \sqrt{9 + 144 + 36} = \sqrt{189} = 3\sqrt{21},
\end{equation*}\]
and therefore the unit vector parallel to \(\overrightarrow{AB}\) is
\[\begin{align*}
&\frac{-3}{3\sqrt{21}} \mathbf{i} + \frac{12}{3\sqrt{21}} \mathbf{j} + \frac{6}{3\sqrt{21}} \mathbf{k} \\
=&\frac{-1}{\sqrt{21}} \mathbf{i} + \frac{4}{\sqrt{21}} \mathbf{j} + \frac{2}{\sqrt{21}} \mathbf{k}.
\end{align*}\]
Find the value of \(p\) such that \(A\), \(B\) and \(C\) are collinear.
The points \(A, B\) and \(C\) are collinear \(\iff \overrightarrow{AB}\) is parallel to \(\overrightarrow{AC} \iff \overrightarrow{AB} = k\overrightarrow{AC}\) for some constant \(k\).
Comparing coefficients, we can say that \(A, B\) and \(C\) are collinear if and only if \(-6 = 3(p-4)\), which is true if and only if \(p = 2\).
Alternatively, we could you use the fact that \(C\) lies on the straight line \(AB\). We could find the equation of the line \(AB\) and then substitute \(C\) into this equation to find \(p\).
If \(p = -2\), find the position vector of \(D\) so that \(ABCD\) is a parallelogram.
