Can Algebraic Calculations Alone Determine Vector Set Constraints Accurately?

[christang_1023]

Homework Statement

Let$\vec{u} = (1, 2)$ and $\vec{v} = (2, 1)$ .

Relevant Equations

Draw the following sets of vectors:
1. ${c\cdot \vec{u}+(1-c)\cdot \vec{v}:c\in R, c\geq0}.$
2. ${a\vec{u}+b\vec{v}:a+b\leq 1}$

1. I consider this problem algebraically, $c\cdot \vec{u}+(1-c)\cdot \vec{v}=c(1,2)+(1-c)(2,1)=(c,2c)+(2-2c,1-c)=(2-c,1+c)$; since the constraint I know is $c\geq 0$, I can conclude the expected vectors$(x,y)$ must have $x\leq2, y\geq 1$.

2. Similarly, I get $a\vec{u}+b\vec{v}=(a+2b,2a+b)$. With the constraint $a+b\leq 1$, since $a,b\in R$, the expected vectors $(x,y)$ should have $x,y\in R$, which means all two-dimensional vectors satisfy the condition.

Am I correct?

 

[fresh_42]

1. Yes, in a way. You do not get all vectors with $x\leq 2 \, , \,y\geq 1$.
You should definitely draw $c\vec{u}+(1-c)\vec{v}$. Insert some values for $c$, e.g. $c\in \{\,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,2,3,4,5\,\}$ and observe what these points have in common.

2. No, not all vectors satisfy the condition. Again, draw the picture for $a+b=1$, then check where $(0,0)$ is, since it is part of the solution set: $a=b=0$. But $(4,5)$ is not.

 

[christang_1023]

1. Yes, in a way. You do not get all vectors with $x\leq 2 \, , \,y\geq 1$.
You should definitely draw $c\vec{u}+(1-c)\vec{v}$. Insert some values for $c$, e.g. $c\in \{\,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,2,3,4,5\,\}$ and observe what these points have in common.

2. No, not all vectors satisfy the condition. Again, draw the picture for $a+b=1$, then check where $(0,0)$ is, since it is part of the solution set: $a=b=0$. But $(4,5)$ is not.

Thank you for your answer. I do observe my answers are not accurate; however, what is wrong with my method, or if there is any modification to make it right?

 

[fresh_42]

Thank you for your answer. I do observe my answers are not accurate; however, what is wrong with my method, or if there is any modification to make it right?

There is nothing wrong doing it algebraically, besides that a) the problem said "draw it", b) you have the wrong description in case 1 and twice as much points as the solution in case 2, and c) that you miss the insights.

Case 1 is a standard construction which is very often used and the clue is, that seeing the equation and having the picture in mind is one and the same thing. It is even worth considering the cases $0\leq c\leq 1$ and the others: $c>1$ or $c<0$ separately.

Case 2 is a region of the plane which often occurs in optimization problems. It is also helpful to see the inequality and automatically associate a geometric object with it. Imagine you had $x \leq 1$. Would this be

... all two-dimensional vectors satisfy the condition

?

 

[christang_1023]

There is nothing wrong doing it algebraically, besides that a) the problem said "draw it", b) you have the wrong description in case 1 and twice as much points as the solution in case 2, and c) that you miss the insights.

Case 1 is a standard construction which is very often used and the clue is, that seeing the equation and having the picture in mind is one and the same thing. It is even worth considering the cases $0\leq c\leq 1$ and the others: $c>1$ or $c<0$ separately.

Case 2 is a region of the plane which often occurs in optimization problems. It is also helpful to see the inequality and automatically associate a geometric object with it. Imagine you had $x \leq 1$. Would this be

?

You are totally right. I made a mistake that the inequality mentioned above cannot express the relationship between $x$ and $y$, which is a significant constraint of $y$.