Linear Algebra Problem: Constraints for Filling a Dashed Triangle

[arpon]

Homework Statement

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

Homework Equations

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, $c+d+e=1;~ 0\leq c ,~ 0\leq d ,~ 0 \leq e;$

 

[SammyS]

Homework Statement

View attachment 101310

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

You have a typo in that combination.

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

Homework Equations

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, $c+d+e=1;~ 0\leq c ,~ 0\leq d ,~ 0 \leq e;$

Those answers are equivalent.

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 .

 

[Ray Vickson]

Homework Statement

View attachment 101310

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

Homework Equations

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, $c+d+e=1;~ 0\leq c ,~ 0\leq d ,~ 0 \leq e;$

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 $c,d,e \leq 1$.

 

[fresh_42]

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