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Math Help - Dot product and unit vectors

  1. #1
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    Dot product and unit vectors

    Hey,

    I am working through work on the dot product and came across this question:

    Let a be a unit vector and let |b| = 4. Let the angle between these two vectors be 30 degrees. Calculate (2a - b) DOT (5a + 3b).

    *Don't know how to make the arrow above the letters, pretend it is above all a's and b's. Also, the DOT is how I am going to say dot product like the formula for the dot product of two vectors with an angle theta between them is: u DOT v = |u| |v| cos(theta).

    I am unsure of how to proceed with this question. I am assuming that I put this through the DOT formula. But am unsure of how to set this up.

    Help?
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  2. #2
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    [tex]\vec{a}\cdot\vec{b}[/tex] gives \vec{a}\cdot\vec{b}.

    (2\vec{a}-\vec{b})\cdot (5\vec{a}+3\vec{b})=10\vec{a}\cdot\vec{a}+\vec{a}\  cdot\vec{b}-3\vec{b}\cdot\vec{b}

    \cos(\theta)=\frac{\vec{a}\cdot\vec{b}}{||a||~||b|  |}.
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  3. #3
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    Let a be a unit vector and let |b| = 4. Let the angle between these two vectors be 30 degrees. Calculate (2a - b) DOT (5a + 3b).

    If a is a unit vector, let it equal {1,0,0}

    then by taking the sin and cos of 30, we know that b would be (\frac{\sqrt{3}}{2},\frac{1}{2},0)

    but since |b|=4, b is really equal to

    4*(\frac{\sqrt{3}}{2},\frac{1}{2},0)=(2\sqrt{3},2,  0)

    now that you know a and b, you should be able to do those dot products pretty easily.
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  4. #4
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    Quote Originally Posted by shenanigans87 View Post
    Let a be a unit vector and let |b| = 4. Let the angle between these two vectors be 30 degrees. Calculate (2a - b) DOT (5a + 3b).

    If a is a unit vector, let it equal {1,0,0}

    then by taking the sin and cos of 30, we know that b would be (\frac{\sqrt{3}}{2},\frac{1}{2},0)

    but since |b|=4, b is really equal to

    4*(\frac{\sqrt{3}}{2},\frac{1}{2},0)=(2\sqrt{3},2,  0)

    now that you know a and b, you should be able to do those dot products pretty easily.
    Okay, using those numbers I get an answer of  2\sqrt{3} - 38

    How did you know to use the coordinates you used for a and b?
    Last edited by Kakariki; April 30th 2010 at 10:15 AM.
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  5. #5
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    All you were told is that a is a unit vector and b is 30 degrees away with a magnitude of 4. You can pick any two vectors that satisfy that, I just picked the easiest.

    And the answer is actually 2\sqrt{3}-38
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  6. #6
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    Quote Originally Posted by shenanigans87 View Post
    All you were told is that a is a unit vector and b is 30 degrees away with a magnitude of 3. You can pick any two vectors that satisfy that, I just picked the easiest.
    Oh! Could you list 2 or 3 others that satisfy the question? Just to be sure that I can understand completely how to choose the vectors for a question like this.

    Thank you so much!
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