Can Algebraic Calculations Alone Determine Vector Set Constraints Accurately?

In summary: Thank you for your explanation.In summary, the conversation discusses two cases where two-dimensional vectors must satisfy certain constraints. In the first case, the constraint is ##c\geq 0## and the expected vectors must have ##x\leq2, y\geq 1##. In the second case, the constraint is ##a+b\leq 1## and all two-dimensional vectors satisfy this condition. However, the method used to arrive at these solutions may not accurately represent the relationship between ##x## and ##y## in the second case.
  • #1
christang_1023
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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?
 
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  • #2
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.
 
  • #3
fresh_42 said:
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?
 
  • #4
christang_1023 said:
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
christang_1023 said:
... all two-dimensional vectors satisfy the condition
?
 
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  • #5
fresh_42 said:
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##.
 
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FAQ: Can Algebraic Calculations Alone Determine Vector Set Constraints Accurately?

1. What is the purpose of drawing sets of vectors?

The purpose of drawing sets of vectors is to visually represent a group of related vectors. This can help with understanding the relationships between the vectors and their properties.

2. How do you draw sets of vectors?

To draw sets of vectors, you can use graph paper or a computer software program. Begin by plotting the coordinates of each vector on the graph paper or inputting the coordinates into the software. Then, use a ruler or the software's drawing tools to connect the points and create the vector. Repeat this process for each vector in the set.

3. Can vectors in a set have different magnitudes and directions?

Yes, vectors in a set can have different magnitudes and directions. This is what distinguishes one vector from another and allows for the representation of different quantities and movements.

4. Are there any rules for drawing sets of vectors?

There are a few rules to keep in mind when drawing sets of vectors. Firstly, the length of the vector should be proportional to its magnitude. Additionally, the direction of the vector should be indicated by an arrow pointing in the direction of its movement. Finally, all vectors in the set should be drawn on the same scale to accurately represent their relationships.

5. How can drawing sets of vectors be useful in scientific research?

Drawing sets of vectors can be useful in scientific research in various ways. It can help with visualizing and analyzing data, identifying patterns and relationships between variables, and making predictions. It is also a useful tool for communication and presenting findings to others.

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