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Math Help - Vectors problems need help

  1. #1
    Junior Member Cyberman86's Avatar
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    Exclamation Vectors problems need help

    my teacher gave me this problems but i just can't figure out this one, please i really need to know this and if you can post Steps on how to do it will be awesome. Also i would rather see the procedure than a lot of explanation.

    Given u=<3,4> w=<-2,2> and v=<-5,6> find the following:

    u + v_____v - u_____4u - 3v_____||4u - 3v||

    Also:

    a unit vector in the direction of v

    find the direction angle for -u

    v*w (is v times w)

    find the angle between v and w

    show that the standard unit vectors i and j are orthogonal.


    I know this long but i hope someone knows this answers, this is not homework to get points, I'm studying for a college test.
    Last edited by Cyberman86; July 28th 2008 at 09:13 AM. Reason: forgot
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  2. #2
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    Quote Originally Posted by Cyberman86 View Post
    Given u=<3,4> w=<-2,2> and v=<-5,6> find the following:

    u + v_____v - u_____4u - 3v_____||4u - 3v||
    u + v = <3,4> + <-5,6> = <3-5,4+6>

    Surely you can do the next one.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Cyberman86 View Post
    my teacher gave me this problems but i just can't figure out this one, please i really need to know this and if you can post Steps on how to do it will be awesome. Also i would rather see the procedure than a lot of explanation.

    Given u=<3,4> w=<-2,2> and v=<-5,6> find the following:

    u + v_____
    Just add the x and y components of each vector together.

    \vec{u}+\vec{v}=\left<3+(-5),4+6\right>=\color{red}\boxed{\left<-2,10\right>}

    v - u_____
    Subtract the corresponding x and y components:

    \vec{u}-\vec{v}=\left<3-(-5),4-6\right>=\color{red}\boxed{\left<8,-2\right>}

    4u - 3v_____
    Multiply each vector by the corresponding scalar coefficient, and then subtract them:

    4\vec{u}-3\vec{v}=\left<4(3)-3(-5),4(4)-3(6)\right>=\color{red}\boxed{\left<27,-2\right>}

    ||4u - 3v||
    Since we have 4\vec{u}-3\vec{v}, apply the equation to find the norm of the vector:

    \parallel 4\vec{u}-3\vec{v}\parallel=\sqrt{27^2+(-2)^2}=\color{red}\boxed{\sqrt{733}}

    Also:

    a unit vector in the direction of v
    first find \parallel\vec{v}\parallel

    \parallel\vec{v}\parallel=\sqrt{(-5)^2+6^2}=\sqrt{61}

    Thus, the unit vector \hat u_v is \color{red}\boxed{\hat u_v=-\frac{5}{\sqrt{61}}\hat i +\frac{6}{\sqrt{61}}\hat j}

    find the direction angle for -u
    I'll get back to this one. Is the angle supposed to be measured with respect to the x axis? I want to say that the angle would be \vartheta=\tan^{-1}\bigg(\frac{u_y}{u_x}\bigg)...

    v*w (is v times w)
    I'm assuming you mean find the dot product?

    When you take the dot product, you're left with a scalar value. You need to multiply the corresponding x and y components together and then add what you have left over:

    \vec{v}\cdot\vec{w}=(-5)(-2)+(6)(2)=\color{red}\boxed{22}

    find the angle between v and w
    The equation for finding the angle between two vectors is this :

    \vartheta=\cos^{-1}\bigg(\frac{\vec{v}\cdot\vec{w}}{\parallel\vec{v  }\parallel\parallel\vec{w}\parallel}\bigg)

    Thus, \vartheta=\cos^{-1}\bigg(\frac{22}{\sqrt{61}\times\sqrt{8}}\bigg)\a  pprox\color{red}\boxed{5.19^{o}}

    show that the standard unit vectors i and j are orthogonal.
    To show that two vectors are orthogonal, you need show that the dot products of the two vectors is zero.

    \hat i \cdot \hat j=\left<1,0\right>\cdot\left<0,1\right>=(1)(0)+(0)  (1)=\color{red}\boxed{0}

    I hope this helps!

    --Chris
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  4. #4
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    I guess we'll never know if Cyberman86 can do them. Oh, well.
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  5. #5
    Junior Member Cyberman86's Avatar
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    Talking

    after seeing Chris showing the process of the Problem i now can do it and its easier than i thought.
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