Understanding Linear Algebra Solutions: Sketching and Ratios Explained

In summary, the sketch doesn't look correct, and the solution is actually on page 6 of the referenced PDF.
  • #1
woundedtiger4
188
0
ifx06a.jpg


Please help me to understand the solution :( (I have made the sketch, is it correct?, where has the 1/4 comes from? is it like 1:3 (1 ratio 3) => numerator divided by numerator + denominator = 1/1+3 =1/4 (as given on page 6 at http://www.lowndes.k12.ga.us/view/14167.pdf )? or am I wrong? For point position vector b in the solution the point B is on the left side from P (as A is on the right side of P)?)

Thanks in advance.
 
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  • #2
woundedtiger4 said:
ifx06a.jpg


Please help me to understand the solution :( (I have made the sketch, is it correct?, where has the 1/4 comes from? is it like 1:3 (1 ratio 3) => numerator divided by numerator + denominator = 1/1+3 =1/4 (as given on page 6 at http://www.lowndes.k12.ga.us/view/14167.pdf )? or am I wrong? For point position vector b in the solution the point B is on the left side from P (as A is on the right side of P)?)

Thanks in advance.
Your sketch isn't very accurate. It looks like A is more than 1/3 of the way along the segment between P and Q. From the problem description, the length of AQ is three times the length of AP, so A should be 1/4 of the way from P to Q.

woundedtiger4 said:
For point position vector b in the solution the point B is on the left side from P (as A is on the right side of P)?)
I don't understand what you're asking here. What point B?
 
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  • #3
woundedtiger4 said:
where has the 1/4 comes from? is it like 1:3 (1 ratio 3) => numerator divided by numerator + denominator = 1/1+3 =1/4 (as given on page 6 at http://www.lowndes.k12.ga.us/view/14167.pdf )? or am I wrong?

It is 1/(1+3) = 1/4. If it were 1/3, then A would be a distance of x from P while A will be a distance of 2x from Q. Can you see this? This would make the ratio 1:2 as opposed to 1:3.

woundedtiger4 said:
For point position vector b in the solution the point B is on the left side from P (as A is on the right side of P)?)
Yes, except that the question doesn't introduce a new point B, it just asked whether A is uniquely determinable, which is no because it can be found on the left side of P as well. A has two possible positions.
 
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  • #4
The line through P and Q is the map ##t\mapsto p+t(q-p)##. Note that this map takes 0 to p and 1 to q. So if we restrict the domain of this map to [0,1], we get the line segment from P to Q. The expression p+t(q-p) can be interpreted as "start at p and take t steps of length |q-p| in the direction towards q". (I know that ##t\leq 1##, but you can certainly imagine taking less than one step; half a step would be a step that's half as long). How big a step would you have to take to end up at a point that's three times as far from P as from Q?

The distinction between P and p is kind of nonsensical if you're working with the vector space ##\mathbb R^3##. It makes sense when you're working with a 3-dimensional affine space, but my guess is that this is from a book that doesn't mention affine spaces.

For the second part, I would find all ##t\in\mathbb R## such that ##|a-p|=3|a-q|##, where ##a=p+t(q-p)=(1-t)p+tq##.
 
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  • #5


I am happy to help you understand the solution to this problem. First of all, it is great that you have made a sketch - visual aids can often help us better understand mathematical concepts.

The 1/4 in the solution refers to the ratio of the length of the segment PB to the length of the segment PA. In other words, it represents the proportion of the distance from P to B compared to the total distance from P to A. This is why it is calculated as 1/(1+3), as you correctly identified.

In terms of the position vector b, it is important to note that position vectors are typically drawn from the origin (0,0) to the point in question. So in this case, the point B is on the left side of P because the vector b starts at (0,0) and ends at B. This is consistent with the sketch provided in the solution.

I hope this helps clarify things for you. Remember, in mathematics, it is always important to carefully define and understand the terminology and symbols used. Keep up the good work in trying to understand the solution and don't hesitate to ask for further clarification if needed.
 

Related to Understanding Linear Algebra Solutions: Sketching and Ratios Explained

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of matrices, vectors, and linear transformations to solve problems in various fields such as engineering, physics, and computer science.

2. What are the applications of linear algebra?

Linear algebra has many applications in different fields such as computer graphics, data analysis, optimization, and machine learning. It is used to solve systems of linear equations, find eigenvalues and eigenvectors, and perform transformations on data sets.

3. What are the basic concepts in linear algebra?

The basic concepts in linear algebra include vector spaces, matrices, linear transformations, determinants, eigenvalues and eigenvectors, and diagonalization. These concepts are used to represent and solve linear equations and systems.

4. What is the difference between a matrix and a vector?

A matrix is a rectangular array of numbers or symbols arranged in rows and columns, while a vector is a one-dimensional array of numbers or symbols. Matrices are used to represent linear transformations, while vectors are used to represent points or directions in space.

5. Can linear algebra be used in real-life situations?

Yes, linear algebra is used in many real-life situations such as calculating the shortest distance between two points, optimizing production processes, predicting stock market trends, and creating computer graphics. It has a wide range of applications in various fields and is an essential tool for problem-solving in the modern world.

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