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Homework Statement
Given ##a\neq b## vectors of ##\mathbb{R}^n##. Determine ##c## which lies in the line segment ##[a,b]=\{a+t(ba) ; t \in [0,1]\}##, such that ##c \perp (ba)##. Conclude that for all ##x \in [a,b]##, with ##x\neq c## it is true that ##c<x##.
Homework Equations
The first part of the question can be solved just using inner product definition.
I don't know how the second part can be solved. But I think could be useful the CauchySchawarz inequality ##<x,y> \leq xy## or maybe the cosine rule...
The Attempt at a Solution
If we suppose that ##c \in [a,b]## then we can write ##c## as:
$$c=a+t_0 (ba)$$
where ##t_0 \in [0,1]##.
By the fact that ##c \perp (ba)## we have:
$$<c,(ba)> \quad = \quad0$$
$$\Rightarrow \quad <a+t_0 (ba),(ba)> \quad = \quad 0$$
$$\Rightarrow \quad <a,(ba)>+<t_0 (ba),(ba)> \quad =\quad 0$$
$$\Rightarrow \quad <a,(ba)>+t_0 ba^2=0$$
$$\Rightarrow t_0 =  \frac{<a,(ba)>}{ba^2} $$
Then ##c## can be uniquely determined as ##c=a+t_0 (ba)##, where ##t_0 =  \frac{<a,(ba)>}{ba^2}##.
$$\ldots$$
Now we need to prove that if we consider any other ##x \neq c## in the line segment ##[a,b]## then ##c<x##.
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