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Math Help - Resolving a vector into a sum of two vectors

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    Resolving a vector into a sum of two vectors

    "Resolve the vector u = 5i + j + 6k into a sum of two vectors, one of which is parallel and the other perpendicular to v = 3i - 6j + 2k."

    What I want to know with this question is how you are able to know how to find the sum of each vector. I know that you find the vector projection of u in the direction of v and the vector component u orthogonal to v for the sum of two vectors. I know how to solve and find the answers from this so, I just basically want to know why you have to use these two in particular to get the answer.
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    Quote Originally Posted by cottontails View Post
    "Resolve the vector u = 5i + j + 6k into a sum of two vectors, one of which is parallel and the other perpendicular to v = 3i - 6j + 2k."

    What I want to know with this question is how you are able to know how to find the sum of each vector. I know that you find the vector projection of u in the direction of v and the vector component u orthogonal to v for the sum of two vectors. I know how to solve and find the answers from this so, I just basically want to know why you have to use these two in particular to get the answer.
    Maybe the picture will well. I can only draw in 2D.

    Resolving a vector into a sum of two vectors-capture.jpg


    First notice the right triangle in red. We need to find the adjacent and opposite sides of it.

    Using trigonometry we know that the adjacent side of a triangle is

    \displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenu  se}}=\frac{A}{H} \iff A=H\cos(\theta)

    The hypotenuse is the magnitude of the vector
    \mathbf{u}

    This gives

    A=||\mathbf{u}||\cos(\theta)

    The object above is a scalar but is has the magnitude of the adjacent, now we just need to point it in the right direction without changing its length.

    So we need a vector pointing in the direction of \mathbf{v} with length 1.

    So lets normalize \mathbf{v}

    Let \displaystyle \mathbf{n}=\frac{\mathbf{v}}{||\mathbf{v}||}

    Now lets combine these two to get

    \displaystyle A\mathbf{n}=(||\mathbf{u}||\cos(\theta))\left(\fra  c{\mathbf{v}}{||\mathbf{v}||} \right)=\frac{||\mathbf{u}||||\mathbf{v}||\cos(\th  eta)}{||\mathbf{v}||^2}\mathbf{v}=\frac{\mathbf{u}  \cdot \mathbf{v}}{||\mathbf{v}||^2}\mathbf{v}

    The last step comes from the geometric definition of the dot product.

    To get the perpendicular \mathbf{p} side

    We use the geometric fact that vectors is Euclidean space add via the parallelogram rule. This gives

    A\mathbf{n}+\mathbf{p}=\mathbf{u} now we just solve for \mathbf{p}, \mathbf{p}=A\mathbf{n}-\mathbf{u}
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