# Linear Algebra Problem

## 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?

## 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;##

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## Answers and Replies

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SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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##

## 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 .

arpon
Ray Vickson
Homework Helper
Dearly Missed

## 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?

## 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##.

arpon
fresh_42
Mentor
If the sum equals 1 and all three summands are positive, how can one be bigger than 1?

arpon