Linear combination and orthogonality

[JulienB]

Homework Statement

Given the vectors a = (5,2,-1), b = (3,2,1), c = (1,2,3), b' = (1,1,0), c' = (3,-3,-2)
We assume that the vector a is a linear combination of the vectors b and c and b' and c' respectively, so that:
a = xb + yc = x'b' + y'c'

a) Determine the factors x and y through construction of different scalar products from the given equation.

b) Repeat the calculation for the vectors b' and c' to obtain x' and y'. What do you notice?

c) How can you verify the conditions given at the beginning?

Homework Equations

Anything about basic vector operations I guess.

The Attempt at a Solution

I think I was able to solve a) and b), but my question is that I have not been using scalar product, but instead a simple system of linear equations:

a) a = xb + yc = (3x,2x,x) + (y,2y,3y) = (3x + y, 2x + 2y, x + 3y)
From there, I determine x and y using the fact that I know the components of the vector a:
3x + y = 5
2x + 2y = 2
x = 2 and y = -1

b) a = x'b' + y'c' = (x',x',0) + (3y',-3y',-2y') = (x' + 3y', x' - 3y', -2y')
Using the same method as a), I obtain:
x' = 7/3 and y' = 1/2

Not only did I not use scalar product (or did I? since multiplying a vector by a constant is considered a scalar product, right?), but I also notice nothing really :)

I would be grateful to read your suggestions.Thank you in advance, I appreciate your help.J.

 

[Mark44]

a) Determine the factors x and y through construction of different scalar products from the given equation.

Scalar product AKA dot product.

Not only did I not use scalar product (or did I? since multiplying a vector by a constant is considered a scalar product, right?),

No. Multiplying a vector by a scalar is considered scalar multiplication. Scalar product implies multiplying two vectors to get a scalar.

This problem is an extension of the work you did in your other thread. Part a is to find constants x and y so that a is a linear combination of b and c. IOW, so that a = xb + yc. xb is the vector projection of a in the direction of b, yc is the vector projection of a in the direction of c. Part b is the same idea, except using two different vectors.

If you do this problem without the use of projections and such, your instructor will almost certainly take off some or all of the points.

 

[JulienB]

Aah yes, I guess I can figure that out now. Thanks a lot again!J.

 

[JulienB]

I keep having a wrong result, can someone tell me where I make a mistake? Thank you very much in advance.

a = xb + yc
xb = ((a.b)/|b|).(b/|b|) = ((a.b)/|b|²).b = (9/7)b
So x = 9/7
yc = ((a.c)/|c|).(c/|c|) = ((a.c)/|c|²).c = (3/7)c
So y = 3/7

However, when I compute the results I found back into xb + yc to check the result, it comes out wrong and is no more equal to a.

 

[JulienB]

What is strange is that it works perfectly in the second part of the problem:

a = x'b' + y'c'
x'b' = ((a.b')/|b'|).(b'/|b'|) = ((a.b')/|b'|2).b' = (7/2)b'
So x' = 9/7
y'c' = ((a.c')/|c'|).(c'/|c'|) = ((a.c')/|c'|2).c' = (1/2)c'
So y' = 3/7

Check: (7/2)b' + (1/2)c' = (7/2,7/2,0) + (3/2,-3/2,-2/2) = (5,2,-1) = a

Did I make (repeatedly!) a calculus error in the first part, or is there something else I didn't quite get? (The last question of b) says "What do you notice?")

 

[Ray Vickson]

What is strange is that it works perfectly in the second part of the problem:

a = x'b' + y'c'
x'b' = ((a.b')/|b'|).(b'/|b'|) = ((a.b')/|b'|2).b' = (7/2)b'
So x' = 9/7
y'c' = ((a.c')/|c'|).(c'/|c'|) = ((a.c')/|c'|2).c' = (1/2)c'
So y' = 3/7

Check: (7/2)b' + (1/2)c' = (7/2,7/2,0) + (3/2,-3/2,-2/2) = (5,2,-1) = a

Did I make (repeatedly!) a calculus error in the first part, or is there something else I didn't quite get? (The last question of b) says "What do you notice?")

The vectors $\vec{b}'$ and $\vec{c}'$ are orthogonal, while $\vec{b}$ and $\vec{c}$ are not.

 

[JulienB]

What I don't get, is how the dot product is even relevant for a) ? With a simple system of linear equations, I come very quickly to the result x = 2 and y = -1, but I can't seem to reach that in any way with the projection of a over b, probably because b and c and indeed not perpendicular.

 

[HallsofIvy]

That's right- the dot product is not relevant in (a). The point is that because b' and c' are orthogonal, it is much easier to find the coefficients to write another vector as a linear combination of them.

If they were orthonormal it would be even easier.

 

[Mark44]

What I don't get, is how the dot product is even relevant for a) ?

It's relevant because its use is a requirement of the problem.

Determine the factors x and y through construction of different scalar products from the given equation.

Is that a good enough reason? Even more important, I believe this problem is setting the stage for a technique to be presented later, the Gram-Schmidt process, which is used to construct an orthogonal basis from a set of basis vectors.

Part a of this problem is to find scalars x and y so that a = xb + yc, with the vectors a, b, and c given.

I believe that the intent of parts a and b is to get you to use the dot product (AKA scalar product) to come up with equations so you can find the constants in each part.

If you dot both sides of the equation above with b, you get another equation. If you dot both sides of the equation above with c, you get another equation. The two equations in x and y can be solved for these variables.

 

[JulienB]

Thank you for your answer. Still, the problem goes:

"a) Determine the factors x and y through construction of different scalar products from the given equation.

b) Repeat the calculation for the vectors b' and c' to obtain x' and y'. What do you notice?"

As Mark said earlier, I don't think I can go through a) without dot product and simply ignore the "repeat the calculation" of b).

 

[Mark44]

Thank you for your answer. Still, the problem goes:

"a) Determine the factors x and y through construction of different scalar products from the given equation.

b) Repeat the calculation for the vectors b' and c' to obtain x' and y'. What do you notice?"

As Mark said earlier, I don't think I can go through a) without dot product and simply ignore the "repeat the calculation" of b).

See my post #9, which talks about how the dot product is intended to be used (I believe).

 

[JulienB]

Thank you Mark, every post you send definitely helps me a lot. The problem was solved using the technique you mentioned in post #9.

 

[Mark44]

Thank you Mark, every post you send definitely helps me a lot. The problem was solved using the technique you mentioned in post #9.

I'm glad to be able to help!