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Math Help - Basic vector question

  1. #1
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    Basic vector question

    I have the problem:

    The scalar projection of a vector v on a vector W is the length of v on w.

    v=3i-4j+k
    w=2i-2j+6

    Now i thought i could just do this:

    (v*w)/|w|

    But then i end up with 24/sqrt(31)
    but the answer is supposed to be 24/7

    I have also tried going the long way by finding the angle and using the cos(x)=a/c to find the length but i end up with the same wrong answer.
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  2. #2
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    Re: Basic vector question

    Quote Originally Posted by mariusg View Post
    The scalar projection of a vector v on a vector W is the length of v on w.
    v=3i-4j+k~\&~ w=2i-2j+6k
    If that is in fact the question then the scalar projection of a vector v on a vector w is:
    \frac{v\cdot w}{\|w\|}=\frac{20}{\sqrt{44}}=\frac{10}{\sqrt{11}  }
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  3. #3
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    Re: Basic vector question

    Thank you.

    I noticed i had made a small typo.

    it was supposed to be w=2i-3j+6k

    As for the question itself, my native language is not english so i have tried to translate it. Something might have gotten "lost in translation" I will try to translate another way:
    Scalar Projection of a vector v on a vector w is the length of v along w. Make use of the dot product to calculate the Scalar Projection in each case.

    But i noticed you used ||w|| insted of |w|. I have not seen that before. Do you mean to take the absolute value of the length of the vector? Isnt that redundant? I thought the length of a vector was always >= 0
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  4. #4
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    Re: Basic vector question

    Quote Originally Posted by mariusg View Post
    Thank you.

    I noticed i had made a small typo.

    it was supposed to be w=2i-3j+6k

    But i noticed you used ||w|| insted of |w|. I have not seen that before. Do you mean to take the absolute value of the length of the vector? Isnt that redundant? I thought the length of a vector was always >= 0
    In most modern textbooks the symbol \|v\| is used to distinguish length of a vector from |\alpha| the absolute value of a sclar. So that \|\alpha v\|=|\alpha|~\|v\|. Take a look at this.

    I know perfectly well what scalar projections are.
    Last edited by Plato; July 27th 2012 at 01:27 PM.
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  5. #5
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    Re: Basic vector question

    Thank you, in my book they have only used single | for both the abs value of the scalar and for the length of a vector. But I see the benefits of using the double || for the length of a vector.

    I apologize if I caused any offence. I only repeated the question exactly as it is written in my textbook(see image and link to Google translate)

    Google translate
    Basic vector question-question.jpg

    Edit:
    My apology is not conditional.
    Last edited by mariusg; July 27th 2012 at 01:43 PM.
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  6. #6
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    Re: Basic vector question

    the english translation is probably something more along the lines of:

    "the (scalar) projection of a vector v on w is the length of v in the direction of w".

    the (vector) projection of v in the direction of w is usually defined to be:

    \frac{v \cdot w}{w \cdot w}w

    as:

    w \cdot w = ||w||^2

    we have that:

    \left|\left|\frac{v \cdot w}{w \cdot w}w\right|\right| = \left|\left|\frac{v \cdot w}{||w||^2}w\right|\right|

    = \left|\frac{v \cdot w}{||w||}\right|\cdot \left|\left|\frac{w}{||w||}\right|\right|

    = \left|\frac{v \cdot w}{||w||}\right| = \frac{|v \cdot w|}{||w||}.

    note that this agrees with Plato's post, up to sign.

    if the angle between v and w is less than (or equal to) a quarter-circle, then the two definitions coincide.

    if the angle between v and w is more than a quarter-circle, then v points "away" from w, and one uses -|v \cdot w| = v \cdot w.

    in other words, Plato's definition is more "accurate" (conveys more geometrical information) than what your book apparently says (real numbers can have a sign, as well as a magnitude), and is in fact, the standard definition.

    in your original problem (as amended), it makes no difference which formula you use, as the angle between (3,-4,1) and (2,-3,6) is indeed less than a quarter-circle.

    so we have that the scalar projection is: (3*2 + (-4)*(-3) + 1*6)/(√[22 + (-3)2 + 62])

    = (6 + 12 + 6)/√49 = 24/7
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  7. #7
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    Re: Basic vector question

    Thank you, you’re explanation has helped me understand the issue a little better Deveno. I had not thought of the sign issue at all.

    The reason I posted my question again was because I had tried exactly what Plato suggested and gotten the wrong answer. When you now showed me that it indeed resulted in the right answer I went back over my work and the logs on my calculator and discovered my error.

    I had entered sqrt[2^2 + (-3)^2 + 6^2] as sqrt[2^2+-3^2+6^2]. When I went over and did all the boring operations manually I discovered that this caused the (-3)^2 to be calculated as -9 instead of 9.

    You learn from your mistakes right?

    Thanks both of you.
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