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Homework Help: Algebra Dot Product question

  1. Nov 28, 2006 #1
    Hi I'm new here and I'm having trouble with this algebra question please help. Sorry if my latex is ugly I'm new to it. I get stuck at the bottom line and I'm not sure how to go further with the question to solve for K

    #7 - Angle between 2 vectors is [tex]\alpha[/tex] where cos[tex]\alpha[/tex] = [tex]\frac{3}{7}[/tex]. a = (2,3,-1) and b = ( -1, K, 1) use the 2 vectors and find possible values for K.

    This is what I did:
    [tex]a\bullet c = |a||b|cos\alpha[/tex]
    (2,3,-1)[tex]\bullet (-1, k, 1) = \sqrt14\sqrt{2+k^2}\frac{3}{7}[/tex]
    [tex]7k-7=\sqrt14\sqrt{2+k^2}[/tex]
     
  2. jcsd
  3. Nov 28, 2006 #2
    Get all the k terms together on 1 side first.
     
    Last edited: Nov 28, 2006
  4. Nov 28, 2006 #3

    dextercioby

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    IF the eq you got is correct, then you can square it already in that form.

    Daniel.
     
  5. Nov 28, 2006 #4
    That would make things measier than need be, because of the -7.
     
  6. Nov 28, 2006 #5
    I actually squared both sides and depending on when you square both sides you'll get different answers. I got 4 answers depending on when I square both sides. I tested the values of K and only 1 out of 4 them are correct, I'll show you what I did.

    [tex]a\bullet b = |a||b|cos\alpha[/tex]
    [tex](2,3,-1)\bullet(-1, K, 1) = (\sqrt{2^2+3^+{-1}^2})(\sqrt{k^2+(-1)^2+1^2})(\frac{3}{7})[/tex]
    [tex]-2 +3k -1 = \sqrt 14\sqrt{k^2+2}(\frac {3}{7})[/tex]
    [tex]3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})[/tex]

    Now you can simplify but multiplying both sides by 7 and dividing by 3 and it will create the line I wrote in the first post
    [tex]7k-7=\sqrt 14\sqrt {k^2+2}[/tex]
    After squaring both sides and solving for K, i got k=-1/5 and =3, both values were incorrect when plugged back into the original equation.

    However if you square both sides on this line one of the answers will be correct:
    [tex]3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})[/tex]
    [tex](3k-3)^2=(\sqrt 14)^2 (\sqrt {k^2+2})^2(\frac {3^2}{7^2})[/tex]
    [tex]9k^2 -18k + 9 = 14 (k^2+2)(\frac {9}{49})[/tex]
    [tex]9k^2 -18k + 9 = 2 (k^2+2)(\frac {9}{7})[/tex]
    [tex]9k^2 -18k + 9 = \frac {18}{7}(k^2+2)[/tex]
    [tex]63k^2 -126k + 63 = 18(k^2+2)[/tex]
    [tex]63k^2 -126k + 63 = 18k^2 + 36[/tex]
    [tex]45k^2 - 126k + 27 = 0 [/tex]
    [tex]9 (5k^2-14k + 3) = 0[/tex]
    [tex]k = \frac {14 +- \sqrt 136}{10}[/tex]
    [tex] K = 2.5662 or k = 1.08377[/tex]

    Only K = 2.5662 actually worked when I tested it:
    [tex]-2 +3k -1 = \sqrt 14\sqrt{k^2+2}(\frac {3}{7})[/tex]
    [tex]\frac {3k-3}{\sqrt 14\sqrt{k^2+2}}=\frac {3}{7}[/tex]
    Pluggin in K = 2.5662
    [tex]0.4286= \frac {3}{7}[/tex]


    http://www.jimloy.com/algebra/square.htm [Broken] -> A link saying squaring both sides can give wrong answers.

    So the question is asking for possible values for K, is there anyway I can solve for more?
     
    Last edited by a moderator: May 2, 2017
  7. Nov 28, 2006 #6
    Show your work for this too: After squaring both sides and solving for K, i got k=-1/5 and =3, both values were incorrect when plugged back into the original equation.

    Try the way I suggested, gather all the K's one one side. And everything else on the other side. See what happens.
     
  8. Nov 28, 2006 #7
    For k=-1/5 and k=3

    [tex]a\bullet b = |a||b|cos\alpha[/tex]
    [tex](2,3,-1)\bullet(-1, K, 1) = (\sqrt{2^2+3^+{-1}^2})(\sqrt{k^2+(-1)^2+1^2})(\frac{3}{7})[/tex]
    [tex]-2 +3k -1 = \sqrt 14\sqrt{k^2+2}(\frac {3}{7})[/tex]
    [tex]3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})[/tex]
    [tex]7k-7=\sqrt 14\sqrt {k^2+2}[/tex]
    [tex]7(k-1)=\sqrt 14\sqrt {k^2+2}[/tex]
    [tex]7^2(k-1)^2=(\sqrt 14)^2(\sqrt {k^2+2})^2[/tex]
    [tex]49(k^2-2k+1) = 14k^2 + 28[/tex]
    [tex]49k^2-98k+49=14k^2+28[/tex]
    [tex]35k^2-98k+21=0[/tex]
    [tex]7(5k^2-14k+3)=0[/tex]
    [tex]7(5k+1)(k-3)=0[/tex]

    K= -1/5 or K=3


    Okay it turns out being really messy so I didn't complete it.

    [tex]3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})[/tex]
    [tex]\frac{3k}{\sqrt{k^2+2}} = \frac {3\sqrt14}{7} + \frac{3}{\sqrt{k^2+2}}[/tex]
    Squaring both sides
    [tex]\frac{9k^2}{k^2+2} = \frac{9(14)}{7} + \frac{9\sqrt14}{7\sqrt{k^2+2}} + \frac{9\sqrt14}{7\sqrt{k^2+2}} + \frac{9}{k^2+2}[/tex]
    Code (Text):
     
     
    Last edited: Nov 28, 2006
  9. Nov 29, 2006 #8

    dextercioby

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    The last line factorization is not correct. So your k=-1/5 and k=3 are not correct. By both methods you'll get the same result, as expected.

    Daniel.
     
  10. Nov 29, 2006 #9
    Why did you backtrack a line? You should have squared it after you multipled through by 7/3, as you did for the other method(s).

    Anyways yeah dextercioby is right. I didn't look too closely. Look closely at your signs. To test it, multiply your factorization through and see what you originally got before you started factorizing.
     
  11. Nov 30, 2006 #10
    Ah I missed that integer sign on the 3. Anyways I went with the answer I got before so thanks for the help guys.
     
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