Linear Algebra Problem

1. May 27, 2016

arpon

1. The problem statement, all variables and given/known data

Under what restrictions on $c, d, e$, will the combinations $c\vec u + d\vec v + e\vec w$ fill in the dashed
triangle?
2. Relevant equations

3. The attempt at a solution
Clearly, $\vec w + a (\vec v - \vec w) + b(\vec u - \vec v)$ will be in the triangle when $0 \leq b \leq a \leq 1$ ;
$\vec w + a (\vec v - \vec w) + b(\vec u - \vec v) = b\vec u + (a-b) \vec v + (1-a)\vec w$
So, we have $c=b, a-b=d, e=1-a$
The restrictions are ,therefore ,
$c+d+e=1$
$0\leq c \leq 1,~ 0\leq d \leq 1,~ 0 \leq e\leq 1$
But according to the book, the restrictions are, $c+d+e=1;~ 0\leq c ,~ 0\leq d ,~ 0 \leq e;$

Last edited: May 27, 2016
2. May 27, 2016

SammyS

Staff Emeritus
You have a typo in that combination.

It should be: $c\,\vec u + d\,\vec v + e\,\vec w$

If c, d, and e are all non-negative and if their sum does not exceed 1, then none of c, d, or e can exceed 1 .

3. May 27, 2016

Ray Vickson

These say the same thing: the constraints $c,d,e \geq 0$, $c+d+e=1$ imply $c,d,e \leq 1$ automatically (think about this). It is harmless but unnecessary to include the explicit upper bounds $c,d,e \leq 1$.

4. May 27, 2016

Staff: Mentor

If the sum equals 1 and all three summands are positive, how can one be bigger than 1?