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Vector and Force Components

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the component of the force [tex]\vec{F} = 2\hat{i} - 2\hat{j} + \hat{k}[/tex] in:

    a) the direction [tex](\hat{i} + \hat{j} - \hat{k})/\sqrt{3}[/tex]

    b) the direction of the vector [tex]3\hat{i} + 2\hat{j} - 6\hat{k}[/tex]

    2. Relevant equations

    [tex]dir \vec{A} = \vec{A}/|\vec{A}|[/tex]

    3. The attempt at a solution

    a) I broke [tex]\vec{F}[/tex] down into components [tex]<2\hat{i}, 2\hat{j}, \hat{k}>[/tex]. Then I figured that F goes [tex]2\hat{i}[/tex] in direction [tex]i[/tex], goes [tex]-2\hat{j}[/tex] in direction [tex]j[/tex], then goes [tex]-\hat{k}[/tex] in direction [tex]k[/tex], giving an overall Force component of [tex]<2\hat{i}, -2\hat{j}, -\hat{k}>[/tex].

    b) I did the same thing and since the signs in each direction were the same as in a), I ended up with the same answer: [tex]<2\hat{i}, -2\hat{j}, -\hat{k}>[/tex].

    It seems to make sense, but I am pretty uncertain, so I thought I would ask here.

    Last edited: Jun 11, 2010
  2. jcsd
  3. Jun 11, 2010 #2


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    Science Advisor

    What you are looking for is the projection of the force vector, [itex]\vec{u}[/itex] on the vector giving the direction, [itex]\vec{v}[/itex]. Imagine a line extended in the direction of the given vector to be projected on and a line segment representing the force vector, having the same beginning point. Dropping a perpendicular to the line from the end of the segment, you have a right triangle with hypotenuse the force vector. Since [itex]cos \theta[/itex] is "near side over hypotenuse", the "near side", the length of the projection, is [itex]|\vec{u}|cos(\theta)[/itex].

    You also need to know that [itex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/itex] so that [itex]\vec{u}|cos(\theta)= \vec{u}\cdot\vec{v}/|\vec{v}|[/itex]. That is the length of the projection. To get it as a vector, multiply by the unit vector in the direction of [itex]\vec{v}[/itex], [itex]\vec{v}/|\vec{v}|[/itex] which gives

    (For the vector in (a), [itex]|\vec{v}|= 1[/itex], for (b), it is 7.)
  4. Jun 14, 2010 #3
    So we're imagining a right triangle in which the given [tex]\vec{F}[/tex] is the adjacent side, and the hypotenuse is the line segment stretching from the origin to the point where a perpendicular from the end of [tex]\vec{F}[/tex] crosses the line formed by an infinite extension of [tex]<\hat{i} + \hat{j} - \hat{k}>/\sqrt{3}[/tex].

    Then you say that the length of the projection is [tex]|\vec{u}|cos\theta[/tex]. What exactly is [tex]\vec{u}[/tex]?
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