# Orthogonal Vectors in Rn Problem

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1. Jul 12, 2018

### Onezimo Cardoso

1. The problem statement, all variables and given/known data
Given $a\neq b$ vectors of $\mathbb{R}^n$. Determine $c$ which lies in the line segment $[a,b]=\{a+t(b-a) ; t \in [0,1]\}$, such that $c \perp (b-a)$. Conclude that for all $x \in [a,b]$, with $x\neq c$ it is true that $|c|<|x|$.

2. Relevant 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 Cauchy-Schawarz inequality $|<x,y>| \leq |x||y|$ or maybe the cosine rule...

3. The attempt at a solution
If we suppose that $c \in [a,b]$ then we can write $c$ as:
$$c=a+t_0 (b-a)$$
where $t_0 \in [0,1]$.
By the fact that $c \perp (b-a)$ we have:
$$<c,(b-a)> \quad = \quad0$$
$$\Rightarrow \quad <a+t_0 (b-a),(b-a)> \quad = \quad 0$$
$$\Rightarrow \quad <a,(b-a)>+<t_0 (b-a),(b-a)> \quad =\quad 0$$
$$\Rightarrow \quad <a,(b-a)>+t_0 |b-a|^2=0$$
$$\Rightarrow t_0 = - \frac{<a,(b-a)>}{|b-a|^2}$$

Then $c$ can be uniquely determined as $c=a+t_0 (b-a)$, where $t_0 = - \frac{<a,(b-a)>}{|b-a|^2}$.
$$\ldots$$
Now we need to prove that if we consider any other $x \neq c$ in the line segment $[a,b]$ then $|c|<|x|$.

Last edited: Jul 12, 2018
2. Jul 12, 2018

### vela

Staff Emeritus
You can show that $x = c + k(b-a)$ for some $k \ne 0$. Then calculate $\langle x,x \rangle$.

3. Jul 12, 2018

### Ray Vickson

The premise of the question is wrong: if $a \neq b, \; a,b \in \mathbb{R}^n$ it is not necessarily true that $c \perp (b-a)$ for some $t \in [0,1].$ For example, if $a = (1,1)$ and $b = (2,1)$, we have $b-a = (1,0)$ and $c = (1,1)+t(1,0) = (1+t,1)$. In order to have $c \perp (b-a)$ we need $t = -1$, so the"perpendicular" $c$ is not on the segment from $a$ to $b$; it is outside that segment, but still on the line through $a$ and $b$.

However, if you allow values $t < 0$ and $t > 1$---in other words, if you allow any $t \in \mathbb{R}$---then the "minimization" result is true. The easiest way to get that is to minimize $F(t) \equiv \| c \|^2$, which is a quadratic function of $t$.

Last edited: Jul 12, 2018
4. Jul 12, 2018

### Onezimo Cardoso

Very well noticed Ray Vickson!
Ok I can reformulate the question as follow:

1. The problem statement, all variables and given/known data
Given $a\neq b$ vectors of $\mathbb{R}^n$. Determine $c$ which lies in the line $r$ determined by in the line segment $[a,b]=\{a+t(b-a) ; t \in [0,1]\}$, such that $c \perp r$. Conclude that for all $x \in r$, with $x\neq c$ it is true that $|c|<|x|$.

The solution I wrote before can be used in the same way, i.e:
$$c = a + t_0 (b-a) \quad where \quad t_0 = - \frac{<a,(b-a)>}{|b-a|^2}$$

But regarding the last part, i.e., to prove that $|x|<|c|$ for all $x \in r$, I have no clue how can I minimize the function $F(t) \equiv \| x \|^2 \equiv \| a+t(b-a) \|^2$.

5. Jul 12, 2018

### Onezimo Cardoso

In fact vela, we can write any $x$ in the line determined by the line segment $[a,b]$ as $x = c + k(b-a)$.
But follow your tip I stucked at the following red question mark:

6. Jul 12, 2018

### vela

Staff Emeritus
Don’t substitute for $c$.

7. Jul 13, 2018

### Ray Vickson

As I said already, that is a quadratic function of $t$. Just expand out $\| a + (b-a)t \|^2$, using the definition of $\| \cdot \|^2.$