= Solution
Take the <cross product> with $\hat{\mathbf q}$:
$$
\mathbf r\times\hat{\mathbf q}
=\mathbf p\times\hat{\mathbf q}
+t(\hat{\mathbf q}\times\hat{\mathbf q})
=\mathbf p\times\hat{\mathbf q}.
$$
Thus one choice is
$$
\boxed{\hat{\mathbf a}=\hat{\mathbf q},\qquad\mathbf c=\mathbf p\times\hat{\mathbf q}}.
$$
Conversely, this <cross product> equation implies $(\mathbf r-\mathbf p)\times\hat{\mathbf q}=0$, so $\mathbf r-\mathbf p$ is parallel to $\hat{\mathbf q}$. It therefore describes precisely the original <straight line>.
Back to article page
