1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Vectors and vector components

  1. Sep 8, 2007 #1
    ^^ ^^
    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

    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.
    Last edited: Sep 8, 2007
  2. jcsd
  3. Sep 8, 2007 #2
    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
  4. Sep 8, 2007 #3


    User Avatar
    Science Advisor

    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?

    ?? 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.

    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.

    Well what "rules" are you talking about? Can you quote them?
  5. Sep 8, 2007 #4
    How do you mean to add the components directly? Can't like terms only be added? i.e. x's with x's etc.
  6. Sep 8, 2007 #5
    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...
  7. Sep 8, 2007 #6


    User Avatar
    Homework Helper

    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..
  8. Sep 8, 2007 #7
    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.
  9. Sep 8, 2007 #8
    Ok, I guess I thought it was more complicated then it was.

    b+c= 2x+y+z

    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:
  10. Sep 8, 2007 #9
    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
    the dot product of u and v =

    so you just multiply the x,y,z components and add them together.
    Last edited: Sep 8, 2007
  11. Sep 8, 2007 #10


    User Avatar
    Homework Helper

    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]
  12. Sep 8, 2007 #11
    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.
  13. Sep 8, 2007 #12


    User Avatar
    Homework Helper

    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...
  14. Sep 8, 2007 #13
    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.
  15. Sep 8, 2007 #14


    User Avatar
    Homework Helper

    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook