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Math Help - Algebraic Vectors

  1. #1
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    Algebraic Vectors

    Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b if |a|=2|b|
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b if |a|=2|b|
    That makes little sense.
    If two vectors are perpendicular the angle between them is \frac{\pi}{2}.

    Are you sure that you asked the right question?
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  3. #3
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    Quote Originally Posted by Plato View Post
    That makes little sense.
    If two vectors are perpendicular the angle between them is \frac{\pi}{2}.

    Are you sure that you asked the right question?
    I will type the question again directly from my textbook without checking the question above.


    Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b, if |a|=2|b|.

    Their answer: 60 degrees.

    They want the angle between vector a and b, not vectors 2a+b and a-3b.
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  4. #4
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    Sorry, I misread the question.
    We get from the given that a\cdot a = 4b\cdot b and 2a\cdot a-5a\cdot b-3b\cdot b=0
    From which it follows that b\cdot b=a\cdot b
    But that means \left(\frac{a\cdot b}{||a||||b||}=\frac{1}{2}\right)
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  5. #5
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    How did you get the a.a=4b.b and 2a.a-5a.b-3b.b=0?
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    Quote Originally Posted by skeske1234 View Post
    How did you get the a.a=4b.b and 2a.a-5a.b-3b.b=0?
    From the given: \left( {2a + b} \right) \bot \left( {a - 3b} \right)\;\& \,\left\| a \right\| = 2\left\| b \right\|
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  7. #7
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    wait, how did you get the -5ab
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  8. #8
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    Quote Originally Posted by skeske1234 View Post
    wait, how did you get the -5ab
    -6a\cdot b+a\cdot b
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  9. #9
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    Quote Originally Posted by skeske1234 View Post
    Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b if |a|=2|b|
    Hell0: this is the solution

    \left( {2a + b} \right) \bot \left( {a - 3b} \right) \Leftrightarrow \left( {2a + b} \right).\left( {a - 3b} \right) = 0<br />
    Calculate :

    \left( {2a + b} \right).\left( {a - 3b} \right) = 0 \Leftrightarrow 2\left\| a \right\|^2 - 6a.b + b.a - 3\left\| b \right\|^2 = 0
    But :

    \left\{ \begin{array}{l}<br />
\left\| a \right\| = 2\left\| b \right\| \\ <br />
a.b = b.a \\ <br />
a.b = \left\| a \right\| \times \left\| b \right\|\cos \theta \\ <br />
\theta = \left( {a,b} \right) \\ <br />
\end{array} \right.<br />
    Subst:
    \left( {2a + b} \right).\left( {a - 3b} \right) = 0 \Leftrightarrow 2\left\| a \right\|^2 - 6a.b + b.a - 3\left\| b \right\|^2 = 0
    Calculate:
    \left( {2a + b} \right).\left( {a - 3b} \right) = 0 \Leftrightarrow 2\left( {2\left\| b \right\|} \right)^2 - 5\left( {2\left\| b \right\|} \right) \times \left\| b \right\|\cos \theta - 3\left\| b \right\|^2 = 0
    I'have :

    \begin{array}{l}<br />
8\left\| b \right\|^2 - 10\left\| b \right\|^2 \cos \theta - 3\left\| b \right\|^2 = 0 \\ <br />
\left\{ \begin{array}{l}<br />
5\left\| b \right\|^2 \left( {1 - 2\cos \theta } \right) = 0 \\ <br />
b \ne \overrightarrow 0 \\ <br />
\end{array} \right. \Leftrightarrow 1 - 2\cos \theta = 0 \\ <br />
\end{array}
    Conclusion:

    \cos \theta = \frac{1}{2} \Leftrightarrow \theta = \frac{\pi }{3}
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