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Finding a unit vector with the given properties

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem states to find a unit vector that is orthogonal to [itex]\left\langle1, 1, -2\right\rangle[/itex], forms an angle of [itex]\frac{\pi}{4}[/itex] with [itex]\left\langle1, 1, 1\right\rangle[/itex] and has v1 > 0.

    2. Relevant equations

    [itex]cos\theta = \frac{\vec{u}\bullet\vec{v}}{|\vec{u}||\vec{v}|}[/itex]

    3. The attempt at a solution

    Since [itex]\vec{v}[/itex] is a unit vector, I know the length is one. This means that in the above equation, the denominator will simply be [itex]|\vec{u}|[/itex].

    I know that [itex]\left\langle1, 1, -2\right\rangle \bullet[/itex] [itex]\left\langle v_{1}, v_{2}, v_{3}\right\rangle = 0[/itex] so [itex]v_{3} = \frac{v_{1}+v_{2}}{2}[/itex]

    [itex]cos(\frac{\pi}{4}) = \frac{\left\langle1, 1, 1\right\rangle \bullet \left\langle v_{1}, v_{2}, v_{3}\right\rangle}{\sqrt{3}} [/itex] so

    [itex]v_{3} = \sqrt{\frac{3}{2}} - v_{1} - v_{2}[/itex]

    I set the two equations for [itex]v_{3}[/itex] equal to each other and end up with [itex]v_{1} + v_{2} = \frac{\sqrt{6}}{3}[/itex] but I'm not sure where to go from here.
     
  2. jcsd
  3. Feb 26, 2013 #2

    Mark44

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    I think you're missing one equation. v is a unit vector, so v12 + v22 + v32 = 1.
     
  4. Feb 26, 2013 #3
    Yes, you're right. I had forgotten about that. Even using that, though, I am still unsure of the next steps I should take. Is the previous work I showed applicable? Was I going in the wrong direction?
     
  5. Feb 26, 2013 #4

    haruspex

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    You now know v3, so you can calculate v32, then v12+v22, etc.
     
  6. Feb 26, 2013 #5

    Mark44

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    You're heading in the right direction. With three equations and three unknowns, you should be able to solve for the coefficients of v.

    You might have a mistake in your work in your second equation.

    cos(##\pi/4##) = √2 /2. I don't see how you got what you show for v3 from your equation.
     
  7. Feb 26, 2013 #6

    haruspex

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    No, I think it's right. More obviously if you write cos(##\pi/4##) = 1/√2
     
  8. Feb 26, 2013 #7
    I think I may be missing something obvious here as I can't seem to figure out how to find v3 from the work I have shown here.
     
  9. Feb 26, 2013 #8

    Mark44

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    No, it's not right.

    You have
    $$\frac{v_1 + v_2 + v_3}{\sqrt{3}} = \frac{1}{\sqrt{2}} $$
    $$ \Rightarrow v_1 + v_2 + v_3 = \frac{\sqrt{3}}{\sqrt{2}}$$

    That's the same as √6/2, not √6/3 as you show.

    Your first two equations are
    ## v_1 + v_2 - 2v_3 = 0##
    ## v_1 + v_2 + v_3 = \frac{\sqrt{3}}{\sqrt{2}}##

    Subtract one of these from the other to solve for v3.

    Then substitute that value in the first equation above to solve for v1 in terms of v2.

    Then substitute for v1 and v3 in the third equation.
     
  10. Feb 26, 2013 #9

    haruspex

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    √6/3 is the value obtained for v1+v2.
     
  11. Feb 26, 2013 #10

    Mark44

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    OK, I didn't catch that.
     
  12. Feb 26, 2013 #11

    haruspex

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    You can't see how to combine those two equations to find v3?
     
  13. Feb 27, 2013 #12
    Ok, so I've found v3 to be [itex]\frac{\sqrt{6}}{6}[/itex].

    I plugged this back into [itex]v_{1} + v_{2} -2v_{3} = 0[/itex] to get [itex]v_{2} = \frac{\sqrt{6}}{3} - v_{1}[/itex].

    I plugged this into [itex]v_{1}^{2} + v_{2}^{2} + v_{3}^2 = 1[/itex] and ended up with [itex]2v_{1}^{2} - \frac{2\sqrt{6}}{3}v_{1} -\frac{1}{6} = 0[/itex].

    From this, I found that [itex]v_{1} = \frac{\sqrt{6}}{6} \pm \frac{1}{2}[/itex]. I choose that [itex]v_{1} = \frac{\sqrt{6}}{6} + \frac{1}{2}[/itex] since it is greater than 0. From here I get that v2 equals the same thing as v1 using what I have above. However, when I use these values for [itex]\vec{v}[/itex], I get that [itex]\vec{v}\bullet\left\langle1, 1, -2\right\rangle = 1[/itex] rather than the 0 that it should be. I can't seem to catch my mistake here.

    EDIT: I've found the mistake. It was a subtraction error. Thank you for the help, everyone.
     
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