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Math Help - Vectors question

  1. #1
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    Vectors question

    I'm not sure where you'd ask vector questions, but since this one involved \cos I decided to ask it here:

    Two vectors, a and b, have a common starting point with an angle of 120 degrees between them. The vectors are such that |a| = 3 and |b| = 4.

    Calculate |a+b|

    The way I thought of it was |a+b| = \sqrt{3^2+4^2-2(3)(4)(\cos{60})} = \sqrt {13}

    However, the answer in the book is \sqrt {3}
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by IanCarney View Post
    I'm not sure where you'd ask vector questions, but since this one involved \cos I decided to ask it here:

    Two vectors, a and b, have a common starting point with an angle of 120 degrees between them. The vectors are such that |a| = 3 and |b| = 4.

    Calculate |a+b|

    The way I thought of it was |a+b| = \sqrt{3^2+4^2-2(3)(4)(\cos{60})} = \sqrt {13}

    However, the answer in the book is \sqrt {3}
    you need to use the dot (inner) product

    |\mathbf{a}+\mathbf{b}|^2=(\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}+\mathbf{b})

    Now expand using the definition of the dot product and remember the geometric definition is

    \mathbf{a}\cdot \mathbf{b}= |\mathbf{a}||\mathbf{b}|\cos(\theta)
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    you need to use the dot (inner) product

    |\mathbf{a}+\mathbf{b}|^2=(\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}+\mathbf{b})

    Now expand using the definition of the dot product and remember the geometric definition is

    \mathbf{a}\cdot \mathbf{b}= |\mathbf{a}||\mathbf{b}|\cos(\theta)
    Sorry, I haven't learned the dot product rule (yet). The way I've been taught so far has just been to "connect" the vectors and use the cosine law.
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  4. #4
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    Quote Originally Posted by IanCarney View Post
    Two vectors, a and b, have a common starting point with an angle of 120 degrees between them. The vectors are such that |a| = 3 and |b| = 4.
    Calculate |a+b|
    This is a result of the parallelogram law.
    If \theta is the angle between \mathbf{\vec{a}}~\&~\mathbf{\vec{b}} then
    |\mathbf{\vec{a}}+\mathbf{\vec{b}}|^2=|\mathbf{a}|  ^2+|\mathbf{\vec{b}}|^2-2|\mathbf{\vec{a}}||\mathbf{\vec{b}}|\cos(\pi -\theta)
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    Quote Originally Posted by Plato View Post
    This is a result of the parallelogram law.
    If \theta is the angle between \mathbf{\vec{a}}~\&~\mathbf{\vec{b}} then
    |\mathbf{\vec{a}}+\mathbf{\vec{b}}|^2=|\mathbf{a}|  ^2+|\mathbf{\vec{b}}|^2-2|\mathbf{\vec{a}}||\mathbf{\vec{b}}|\cos(\pi -\theta)
    Would you agree with my answer then? |a+b| =
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  6. #6
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    Quote Originally Posted by IanCarney View Post
    Would you agree with my answer then? |a+b| =
    Draw a sketch to scale for clarification (see attachment)

    ... and \sqrt{13} \approx 3.60555

    It seems to me that there is a typo in your book.
    Attached Files Attached Files
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  7. #7
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    Another way to do this is to note that you have two sides of a triangle, having lengths 3 and 4 with angle between them 180- 120= 60 degrees. The length of the third side, which is |a+ b| can be calculated using the cosine law.
    \sqrt{13}
    is correct.
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  8. #8
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    Quote Originally Posted by earboth View Post
    Draw a sketch to scale for clarification (see attachment)

    ... and \sqrt{13} \approx 3.60555

    It seems to me that there is a typo in your book.
    Thanks! I had done that but the answer in the book threw me off.

    Another question: Calculate the angle between vector a and vector a+b.

    I would do sin60/sqrt{13}=sinx/4, where x is the angle, right? The answer in the book appears to be another typo, so I just wanted some clarification.
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    Quote Originally Posted by IanCarney View Post
    Thanks! I had done that but the answer in the book threw me off.

    Another question: Calculate the angle between vector a and vector a+b.

    I would do sin60/sqrt{13}=sinx/4, where x is the angle, right? The answer in the book appears to be another typo, so I just wanted some clarification.
    You have three sides and one angle of a triangle. Use Law of Sines. What answer did you get?

    -Dan
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  10. #10
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    Quote Originally Posted by IanCarney View Post
    Thanks! I had done that but the answer in the book threw me off.

    Another question: Calculate the angle between vector a and vector a+b.

    I would do sin60/sqrt{13}=sinx/4, where x is the angle, right? The answer in the book appears to be another typo, so I just wanted some clarification.
    1. Compare my reply (#6):
    \vec r = \vec a + \vec b
    \angle(\vec a, \vec r) = \beta

    2. Y0u know:

    |\vec a| = 3 , |\vec b| = 4 , |\vec r| = \sqrt{13}

    3. According to the Cosine rule you have:

    (\vec b)^2= (\vec a)^2+(\vec r)^2 - 2 \cdot \vec a \cdot \vec r \cdot \cos(\beta)

    Determine \cos(\beta) and consequently .
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  11. #11
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    Quote Originally Posted by earboth View Post
    1. Compare my reply (#6):
    \vec r = \vec a + \vec b
    \angle(\vec a, \vec r) = \beta

    2. Y0u know:

    |\vec a| = 3 , |\vec b| = 4 , |\vec r| = \sqrt{13}

    3. According to the Cosine rule you have:

    (\vec b)^2= (\vec a)^2+(\vec r)^2 - 2 \cdot \vec a \cdot \vec r \cdot \cos(\beta)

    Determine \cos(\beta) and consequently .
    Yeah, you can do it that way too. (I just like the Law of Sines better in this circumstance. But no worries.)

    -Dan
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