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Homework Help: Need help with this Vector Question! +rep

  1. Nov 10, 2009 #1
    Alright all im on this vector question and im having problems with this question. Part (i) i have no idea what to start with / how i possibly prove the question.

    This is the question, help will be deeply appriciated. Thanks, David. Also note that
    || a || = (a_1 + a_2 + ... + 1_n)^1/2 .

    || a - b || = || a || || b || - 2 (a \cdot b).

    I have done part (i) :) Its the rest im struggling with.

    Last edited: Nov 10, 2009
  2. jcsd
  3. Nov 10, 2009 #2


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    Staff: Mentor

    Could you show us how you solved part (i)?
  4. Nov 10, 2009 #3
    c = a + t(b-a), take t = l1 / l
    c = a + (l1 / l)(b-a), Expand
    c = a -a(l1 / l) + b(l1 / l) Collect terms with a.
    c = a(1 - (l1 / l)) + b(l1 / l) 1 = l / l
    c = a(l/l - (l1 / l)) + b(l1 / l)
    c = a(l - l1)/l + b(l1 / l) l - l1 = l2
    c = a(l2 / l) + b(l1 / l)

    Its the rest im stuck on :(
    Any ideas?
  5. Nov 10, 2009 #4


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    Hi simpleas123, welcome to PF!:smile:

    You should probably use boldface type to distinguish vectors from scalars, just to make things clear (Clicking on the "go advanced" button will give you all kinds of typesetting options, including using [itex]\LaTeX[/itex])

    Also, although [itex]t=\frac{l_1}{l}[/itex] is correct, you need to provide a justification of this for your proof to be complete (A justification of the hint given is also necessary).

    For problem (ii), proving the statement "If [itex]\textbf{x}=\textbf{y}[/itex], then [itex]\textbf{c}\cdot\textbf{x}=\textbf{c}\cdot\textbf{y}[/itex] and [itex]\textbf{c}\times\textbf{x}=\textbf{c}\times\textbf{y}[/itex]" should be easy enough...as for proving the opposite direction if-then statement; what is the result of problem 3 (b) they mention?....What is property (4) of the cross product?
  6. Nov 10, 2009 #5
    Property 4 is the same as result of problem 3 (b).
    It states that;

    Where a,b are vectors

    a x b = 0, if and only if there exist two scalars [tex]\mu,\lambda[/tex] one of which is non-zero such that a[tex]\lambda[/tex] = b[tex]\mu[/tex]

    I have done part ii) now too.
  7. Nov 10, 2009 #6


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    Okay, and what was your solution for (ii)?

    Problem (iii) appears to contain a typo...I am fairly certain it's suppose to say:

    "Show that the equation of the line [itex]l[/itex] through the points [itex]\textbf{x}_0[/itex] and [itex]\textbf{x}_0+\textbf{a}[/itex] is [itex](\textbf{x}-\textbf{x}_0)\times\textbf{a}=0[/itex]"

    To do this, just use what property 4 tells you about the equation [itex](\textbf{x}-\textbf{x}_0)\times\textbf{a}=0[/itex]...
  8. Nov 10, 2009 #7
    Part ii)
    [itex]c \cdot y = c \cdot x[/itex]
    [itex](c \cdot y) - (c \cdot x) = 0[/itex]
    [itex]c \cdot (x-y) = 0[/itex]

    [itex]\therefore y = x[/itex]
    and it was pretty much the same method for the cross product execpt u had to state a property at the end as x x y could equal something other than 0.
  9. Nov 10, 2009 #8


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    Careful, [itex]\textbf{c}\cdot(\textbf{x}-\textbf{y})=0[/itex] doesn't necessarily mean that [itex]\textbf{x}-\textbf{y}=0[/itex]...one of the vectors in a dot product being zero is not the only way for the dot product to be zero...
  10. Nov 10, 2009 #9


    Staff: Mentor

    [itex]\bold{c} \cdot (\bold{x}-\bold{y}) = \bold{0}[/itex]
    This doesn't necessarily imply that x = y. All it says is that c is orthogonal to x - y.

    Counterexample: c = (1, 0, 0), x = (0, 2, 0), y = (0, 1, 0).
    [itex]\bold{c} \cdot (\bold{x}-\bold{y}) = (1, 0, 0) \cdot (0, 1, 0) = \bold{0}[/itex]
    but x - y isn't 0.
  11. Nov 10, 2009 #10
    Oh snap thats right, well then how would i prove it?
  12. Nov 10, 2009 #11


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    Well, [itex]\textbf{c}\cdot(\textbf{x}-\textbf{y})=0[/itex] tells you either [itex]\textbf{x}-\textbf{y}=0[/itex] or [itex]\textbf{x}-\textbf{y}[/itex] is orthogonal to [itex]\textbf{c}[/itex]...right?

    What can you say about [itex]\textbf{c}\times(\textbf{x}-\textbf{y})[/itex] if [itex]\textbf{x}-\textbf{y}[/itex] is orthogonal to [itex]\textbf{c}[/itex]?
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