Further assume \(\vec {p}'(0) = \vector {1,1,1}\) and \(\vec {q}'(0) = \vector {1,2,3}\).
Find real numbers \(a = \answer {2}\) and \(b = \answer {-1}\) so that the vector-valued function
\[ \vec {r}(t) = a \cdot \vec {p}(t) + b\cdot \vec {q}(t) \]
satisfies \(\vec {r}'(0) = \vector {1,0,-1}\).
By the constant multiple and sum rules for vector-valued derivative, \(\vec {r}'(t) = a \cdot \vec {p}'(t) + b \cdot \vec {q}'(t)\).
Consequently, we want \(\vec {r}'(0) = \vector {1,0,-1}\) to equal \(a \cdot \vec {p}'(0) + b \cdot \vec {q}'(0)\).
But \(a \cdot \vec {p}'(0) + b \cdot \vec {q}'(0)\) equals \(a \cdot \vector {1,1,1} + b \cdot \vector {1,2,3}\).
So we wish to solve \(a \cdot \vector {1,1,1} + b \cdot \vector {1,2,3} = \vector {1,0,-1}\).
This is a system of three equations in two unknowns, namely that \(a + b = 1\) and \(a + 2b = 0\) and \(a + 3b = -1\).
Subtracting the second equation from the first reveals that \(b = -1\).
Then since \(a + b = -1\), it must be that \(a = 2\).