Scale two vectors to match third one

Question

Suppose three vectors $a, b, c \in \mathbb{R}^l$ are given. What would be the best approach to scale $b$ and $c$ (with numbers $s, t \in \mathbb{R}$ respectively) in such a way that the distance $a-(sb+tc)$ gets minimized (optimal case: $a=tb+sc$).
I've thought about using the first two moments for calculating the parameters: $E[a]=E[sb+tc]=sE[b]+tE[c]$ and $Var[a]=Var[sb+tc]=s^2Var[b]+t^2Var[c]$. Would that make for a good approximation?

Answer 1

Hint: study the polynomial: $$ f(s,t) = \|a − (sb + tc)\|^2 = $$ $$ = a\cdot a - 2(a\cdot b)s - 2(a\cdot c)t + 2(b\cdot c)st + (b\cdot b)s^2 + (c\cdot c)t^2. $$ $$0 = \frac{\partial f}{\partial s} = \cdots$$ $$0 = \frac{\partial f}{\partial t} = \cdots$$

Answer 2

The case $a=tb+sc$ is achievable only if $a,$ $b,$ and $c$ are linearly dependent; since they are three vectors, that would mean they all lie in a single subspace of dimension $2,$ that is, in the same two-dimensional plane. If $l > 2$ it is not guaranteed that three arbitrary vectors will all lie in one plane.

Your use of notations such as $E[a]$ and $Var[a]$ implies that you think the vectors are random. But if $E[a],$ $E[b],$ and $E[c]$ exist, each is a vector in $\mathbb R^l,$ hence you have the same problem solving $E[a]=sE[b]+tE[c]$ as you do for solving $a=sb+tc$ when $a,$ $b,$ and $c$ are known vectors: there is a solution only in the special case in which the three vectors lie in one subspace of dimension $2.$

In the case where you cannot set $a - (sb+tc) = 0$ exactly, you can minimize the difference by projecting $a$ orthogonally onto the plane spanned by $b$ and $c.$ Since the projected vector $a'$ is in the same plane as $b$ and $c,$ it is linearly dependent on them, and you can solve $a' = sb+tc.$