96fbd5…: “Close to the bad equation”

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Suppose $\vec {p} : \R \to \R ^3$ and $\vec {q} : \R \to \R ^3$ are vector-valued functions.

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 $a = 2$ .