Vectors and vector components

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  • #1
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^^ ^^
1.) Given the two vectors b=x+y and c=x+z find b+c, 5b+2c



2.) r+s = (r1+s1, r2+s2, r3+s3), if r and s are vectors with components, r=(r1,r2,r3) and
s=(s1,s2,s3)



3.) Is this correct for b+c?
b+c= (bx+cx,by,cz)



To me that doesn't look right, but I think I'm applying the rules correctly. I don't know how to approach 5b+2c because I don't know now to apply the scalars 5 and 2 to the vector components. How is that subquestion approached?

thanks alot

Edit: the ^ signs are supposed to appear over the x+y and the x+z within the b and c vectors.
 
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Answers and Replies

  • #2
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no that's not right.
you don't need to multiply the components x,y by b. Just add the components directly.

When you do 5b+2c first figure out what 5b is, just multiply each component by 5. The same with 2c, and then add the components like you do for b+c
 
  • #3
HallsofIvy
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^^ ^^
1.) Given the two vectors b=x+y and c=x+z find b+c, 5b+2c
Surely, this is next best thing to trivial! Ignore the fact that these are vectors.
If these were just algebraic expressions and you are told that b= x+y, c= x+ z, what would you get substituting those into b+ c and 5b+ 2c?

2.) r+s = (r1+s1, r2+s2, r3+s3), if r and s are vectors with components, r=(r1,r2,r3) and
s=(s1,s2,s3)
?? I don't see a question here!
Are you asking if, given that r= (r1,r2,r3) and s= (s1,s2,s3), is r+s= (r1+s1, r2+s2, r3+s3)? It certainly is! That's the definition of addition of vectors in component form.

3.) Is this correct for b+c?
b+c= (bx+cx,by,cz)
Well, it can be if cy= 0 and bz= 0. In general, if b= (bx,by,bz), c= (cx,cy,cz) then
b+ c= (bx+cx,by+cy,bz+cz). If it happens that b= (bx,by,0), c= (cx,0,cz) then
b+c= (bx+cx,by,cz)- but only in that case.



To me that doesn't look right, but I think I'm applying the rules correctly. I don't know how to approach 5b+2c because I don't know now to apply the scalars 5 and 2 to the vector components. How is that subquestion approached?

thanks alot

Edit: the ^ signs are supposed to appear over the x+y and the x+z within the b and c vectors.
Well what "rules" are you talking about? Can you quote them?
 
  • #4
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How do you mean to add the components directly? Can't like terms only be added? i.e. x's with x's etc.
 
  • #5
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Halls of ivy, i was given a template to follow and 2.) was 'any given formulas'; that is what I was given. You are correct in saying that it is a definition.

So my first question is wrong your saying? cy wasn't even given in the question...
 
  • #6
learningphysics
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As Halls said, just treat them like ordinary algebra variables. They want you to get b+c and 5b+2c... in terms of x and y. How would you do it if they weren't vectors but just normal variables..
 
  • #7
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changing the form of the vectors a bit might help you see how to add them:
b=1x+1y+0z, c=1x+0y+1z
so for b+c=? just replace b by it's equation and c by it's equation and add.
 
  • #8
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Ok, I guess I thought it was more complicated then it was.

b+c= 2x+y+z
5b+2c=7x+5x+2z

the next part was to do bc as a dot product, the formula for this is AB=ABcos(theta)
since no angle is given, the answer is simply:
x^2+xz+yx+yz
ya?
 
  • #9
492
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no that's not right it's actually that: ||A||*||B||*cos(theta) = A dot B
where ||A|| = distance of A to the origin: [itex]\sqrt{x^2+y^2+z^2}[/itex]

you can use the definition that if you have
[tex]\vec{u}=u_x+u_y+u_z[/tex]
[tex]\vec{v}=v_x+v_y+v_z[/tex]
the dot product of u and v =
[tex]u_x*v_x+u_y*v_y+u_z*v_z[/tex]

so you just multiply the x,y,z components and add them together.
 
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  • #10
learningphysics
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Ok, I guess I thought it was more complicated then it was.

b+c= 2x+y+z
5b+2c=7x+5x+2z

the next part was to do bc as a dot product, the formula for this is AB=ABcos(theta)
since no angle is given, the answer is simply:
x^2+xz+yx+yz
ya?
Your answer is correct, though be careful about the right notation and all that...

So this would work:
[tex]\vec{x}\cdot\vec{x} + \vec{x}\cdot\vec{z} + \vec{y}\cdot\vec{x} + \vec{y}\cdot\vec{z}[/tex]

or this one:

[tex]|\vec{x}|^2 + \vec{x}\cdot\vec{z} + \vec{y}\cdot\vec{x} + \vec{y}\cdot\vec{z}[/tex]

Because the angle between [tex]\vec{x}[/tex] and itself is 0, so the cosine of 0 is 1, and the dot product of x with itself becomes [tex]|\vec{x}|^2[/tex]
 
  • #11
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But in A dot B you need to know the angle between A and B, in this case the angle between the vector's x+y and x+z.

Usually you should use A dot B = ||A||||B||cos(theta) when the vectors are given with numbers and an angle.
 
  • #12
learningphysics
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But in A dot B you need to know the angle between A and B, in this case the angle between the vector's x+y and x+z.

Usually you should use A dot B = ||A||||B||cos(theta) when the vectors are given with numbers and an angle.
Yes, if he knows the angles he should use them. I assumed the angles weren't given. In which case, I don't see what's wrong with just writing [tex]\vec{A}\cdot\vec{B}[/tex]

I'm assuming x,y and z are arbitrary vectors... not necessarily in the x direction y direction or z direction. The question doesn't specify them being any specific type of vectors...
 
  • #13
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hm..well using the other definition of dot product I get x^2 as the solution since (x*x+y*0y+0z*z)=x^2

and I think that book uses x/y/z instead of i/j/k but if it's wrong it should still be x^2 I think.
 
  • #14
learningphysics
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hm..well using the other definition of dot product I get x^2 as the solution since (x*x+y*0y+0z*z)=x^2

and I think that book uses x/y/z instead of i/j/k but if it's wrong it should still be x^2 I think.
If they are arbitrary vectors then it won't be x^2.

But I think you're right about them being i/j/k, since he mentions the "^" in which case the answer is 1. |i|^2 = 1.

Sorry if I confused the OP. :redface:
 

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