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How do I begin to solve this vector/cross product problem?

  1. Oct 11, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    It is given that the soltuion of the vector equation y x a=b is

    [tex]\underline{y}= \lambda \underline{a} + \frac{\underline{a} \times \underline{b}}{| \underline{a}|^2}[/tex]

    with a . b=0 and [itex]\lambda[/itex] is a scalar. Use this information to find the solution of the equation (x x a) + (x . b)c=d.

    Where x is the unknown vector and [itex]\underline{a} \cdot \underline{c} \neq 0[/tex]


    2. Relevant equations

    [tex]\underline{A} \cdot \underline{B}= |\underline{A}| |\underline{B}|cos\theta[/tex]

    [tex]\underline{A} \times \underline{B} = |\underline{A}| |\underline{B}|sin\theta \hat{n}[/tex]


    3. The attempt at a solution

    This is my method of thinking.

    If [itex]\underline{a} \cdot \underline{b} =0 [/itex] then this means that a and b are perpendicular

    [tex]y \times a =b[/tex]

    [tex]a \time (y \times a)= a \times b[/tex]

    [tex]= y(a \cdot c)-a(a \cdot y)= a \times b[/tex]

    [tex]\Rightarrow = y(a \cdot c)=a(a \cdot y)+a \times b[/tex]

    [tex]\div a \cdot c[/tex]

    [tex]y= a \frac{a \cdot y}{a \cdot c} + \frac{a \times b}{a \cdot c}[/tex]

    Comparing this with the given solution:

    [tex]\lambda = \frac{a \cdot y}{a \cdot c}[/tex]

    AND

    [tex]|a|^2={a \cdot c}[/tex]

    On the right track so far?


    [tex](x \times a) + (x \cdot b)c=d[/tex]

    [tex]a \times (x \times a)+ a \times (x \cdot b)c= a \times d[/tex]

    [tex] x(a \cdot a)+ a \times c(x \cdot b)=a \times d[/tex]

    [tex]x |a|^2 + a \times c(x \cdot b)= a \times d[/tex]

    [tex]x(a \cdot c) + a \times c( x \cdot b)= a \times d[/tex]

    and I am stuck here.
     
  2. jcsd
  3. Oct 11, 2008 #2

    D H

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    Correct so far.

    OK, but you might want to expand [itex]y\times a[/itex] first. What is [itex]\lambda a \times a[/itex]?

    Whoa!! Where did that c come from?

    Hint: You can't solve for lambda. Its value is completely indeterminate here.
     
  4. Oct 11, 2008 #3

    rock.freak667

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    What do you mean by expanding [itex]y\times a[/itex] ? wouldn't [itex]\lambda a \times a[/itex] just be 0 since a x a =0?

    ah that came from whenever I expanded a x (y x a), I wrote above it a x (b x c)= b(a.c)-c(a.b) and then forgot to replace the letters...dumb me...

    but this is the correct thing, I hope:

    [tex]y=\frac{a \times b}{|a|^2} + a \frac{a \cdot y}{|a|^2}[/tex]
     
    Last edited: Oct 11, 2008
  5. Oct 11, 2008 #4

    D H

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    Exactly. The given equation yields absolutely no information about lambda. It can take on any value; you cannot solve for it.

    Given that [itex]a \cdot b = 0[/itex], the equation for [itex]y[/itex] is an identity. You are supposed to use this identity to help solve the problem.
     
  6. Oct 11, 2008 #5

    rock.freak667

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    Well I do not see how I could use that. Unless I take [itex](x \times a) + (x \cdot b)c=d
    [/itex] and then cross all of that with b
     
  7. Oct 11, 2008 #6

    D H

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    Try taking the inner product with a instead.
     
  8. Oct 11, 2008 #7

    rock.freak667

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    Then I should get:

    [tex] a \cdot (x \times a) + a \cdot c (x \cdot b)=a \cdot d[/tex]

    [tex]x \cdot (a \times a)+ a \cdot c (x \cdot b)=a \cdot d[/tex]

    Since [itex] a \times a=0[/tex]

    [tex] x \cdot b=\frac{a \cdot d}{a \cdot c}[/tex]


    I do not think I can isolate x from this.
     
  9. Oct 11, 2008 #8

    D H

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    So put this result back into your original equation for x. Can you use the given identity yet? If not, what can you do to make that identity usable?

    BTW, you should now see why the problem statement asserts that [itex]a\cdot c \ne 0[/itex].
     
  10. Oct 11, 2008 #9

    rock.freak667

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    If I put it back into the original problem, I will get:

    [tex](x \times a) + \frac{a \cdot d}{a \cdot c} c= d[/tex]

    [tex]\Rightarrow x \times a = t \ Where \ t=d-\frac{a \cdot d}{a \cdot c} c[/tex]

    [tex]\Rightarrow x=\lambda a + \frac{a \times t}{|a|^2}[/tex]
     
  11. Oct 11, 2008 #10

    D H

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    Excellent.
     
  12. Oct 11, 2008 #11

    rock.freak667

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    Thanks!


    (I just wasted a bunch of time solving y x a =b though :biggrin:)
     
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