Write a vector as the combination of 2 other vectors

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SUMMARY

The discussion focuses on decomposing the vector u=(2,3,-1) into two components: one parallel to v=(1,0,-3) and the other orthogonal to v. Participants emphasize the importance of using the dot product to ensure orthogonality, where w·v=0, and the projection of u onto v to find the parallel component. The correct approach involves setting up the equations kv + w = u, ensuring w is perpendicular to kv, and applying the Pythagorean theorem to relate the lengths of the vectors. The solution requires careful organization of information and understanding of vector geometry.

PREREQUISITES
  • Understanding of vector decomposition and representation
  • Familiarity with dot product and its implications for orthogonality
  • Knowledge of vector projection techniques
  • Basic principles of geometry related to triangles and vector addition
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  • Study vector projection methods, specifically how to calculate the projection of one vector onto another.
  • Learn about the geometric interpretation of vector addition and the Pythagorean theorem in vector spaces.
  • Explore the concept of orthogonal vectors and their applications in linear algebra.
  • Practice solving systems of equations involving vectors to solidify understanding of vector relationships.
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Students studying linear algebra, particularly those focusing on vector decomposition, geometry, and applications of the dot product. This discussion is also beneficial for educators seeking to clarify vector concepts for their students.

MarcL
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Homework Statement


Write the vector u=(2,3,-1) as the sum of two vectors, one parallel to v=(1,0,-3) and the other orthogonal to v=(1,0,-3)

Homework Equations


orthogonal vector would imply that v"dot"w = 0 ( dot product)

Parallel vector would imply that v=kz
OR the crossproduct of both is 0

The Attempt at a Solution



My attempt went as follow:
I defined 2 vectors that w+z = u where u=(2,3,-1)

Vector w is orthogonal to v and therefore --> w"dot"v = 0 --> w1*1+w2*0+w3*-3=0

And that z= kv which would give z= (1*k,0*k,3*k)

I thought that I could define any vector orthogonal that would give 0, for instance (3,1,1) would give 0.

However the system always ends up being inconsistent, I don't know where I might be going wrong :/
 
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For u = av + bw then w must be in the same plane as u and v and in addition it must be perpendicular to v as stated in your problem. The a and b are scalars.

Try drawing a picture and see if you can figure out w from the geometry and your vectors.
 
two questions from this: is it obligatory to use a scalar in that equation for the w?

2) I don't see the point in understanding that w must be in the same plane as u and v, I mean I don't see how it helps me solve the problem.

Thing is, I chose an arbitrary vector for w that would satisfy the orthogonal requirement. However, using it, I get an a=-7 and b=3 which gives me a z parameter that is wrong :/
 
There is a whole plane of vectors that are orthogonal to v but you must choose the one that is in the same plane as u and v right?

Try to visualize it and you'll see all three vectors must be in the same plane.

Another way to think of it is vector addition is a parallelogram with v and w as sides and with the diagonal between them being u the sum of the vectors.
 
Is it kinda like spanning? I mean two vectors that add up to any vector in a plane?
 
but how would I go then to find my problem, for spanning I usually use a matrix to find my coefficient, but I don't know if that would apply in this case as it has to satisfy requirements, no?
 
I'm thinking you should look at u - k.v = w and v.w=0
 
For this problem you need to find the vector projection of u in the direction of v, which is denoted in some texts as ##\vec{Proj_v u}##. If you've been assigned this problem, it's a fairly safe bet that this concept has been presented in class. Once you have that vector, find a vector perpendicular to the vector projection, so that the two vectors you found add vectorially to u.

Draw a sketch. The two vectors you're looking for are the sides of a right triangle. Forget about matrices and spanning sets and focus on the geometry here.
 
  • #10
Well I can see where the projection comes from, how you use it geometrically. Only thing is that , in class, my teacher used a set of equation from the orthogonal and parallel requirement, so I assumed I had to do it that way, and that way I don't understand.
 
  • #11
You have u and you have the project of u on v so you have the lengths of two sides of a triangle with w being the third side.
 
  • #12
I realize I can do it with the projection, but we didn't use projection for that in class, so I believe my teacher would prefer if we solved it with system of equation(s). This is where I get confused, I don't know hjow to get it that way.
 
  • #13
I'd do it the way you know how and ask the teacher about the other method later. You might discover that that method doesn't even apply in this case.
 
  • #14
MarcL said:

Homework Statement


Write the vector u=(2,3,-1) as the sum of two vectors, one parallel to v=(1,0,-3) and the other orthogonal to v=(1,0,-3)

Homework Equations


orthogonal vector would imply that v"dot"w = 0 ( dot product)

Parallel vector would imply that v=kz
OR the crossproduct of both is 0

The Attempt at a Solution



My attempt went as follow:
I defined 2 vectors that w+z = u where u=(2,3,-1)

Vector w is orthogonal to v and therefore --> w"dot"v = 0 --> w1*1+w2*0+w3*-3=0

And that z= kv which would give z= (1*k,0*k,3*k)
You wrote the equation u=w+z. Multiplying it with v ( dot product) you get ##\vec v \cdot \vec u = \vec v \cdot \vec w + \vec v \cdot \vec z ##. The first product is zero. z=kv. Find k.
 
  • #15
The problem works out pretty nicely if you organize your information. You're given vector u and v. The vectors you want are kv and w.

The three constraints yield a consistent set of equations so that you can find the scalar k and the coordinates of w.

1. kv + w = u
2. kv is perpendicular to w
3. Pythagorean theorem on the lengths of kv, w, and u

BTW, the method I mentioned earlier will also work, but as it hasn't been presented yet, probably shouldn't be used.
 
  • #16
Mark44 said:
The problem works out pretty nicely if you organize your information.
Hard to believe there's a simpler way than ehild's above.
 
  • #17
I just did this ( realized where my mistake was)

(2,3,-1)=a(1,0,3)+b(3,c,1) where the second vector comes from w1*1+)-3W3=0 --> W1=3W3 so I can just say 3,c,1,

Solve for 3 variables with 3 equation. I just didn't use the c but took any number at first, without understanding why ^^
 

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