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Math Help - Perpendicular Unit Vectors

  1. #1
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    Perpendicular Unit Vectors

    For this question, I already have the correct answer of t = 3. My question is, do we ignore \vec{j}(4t)=-2\vec{j}\rightarrow t=-2 because the number can only be 0 where t\vec{v}+\vec{u} lies on the \vec{j}-axis?

    Q.
    If \vec{u}=3\vec{i}+2\vec{j} & \vec{v}=-\vec{i}+4\vec{j} & if t\vec{v}+\vec{u} is on the \vec{j}-axis, find the value of the scalar t.

    Attempt: t(-\vec{i}+4\vec{j})+3\vec{i}+2\vec{j}=0
    -t\vec{i}+4t\vec{j}=-3\vec{i}-2\vec{j}
    \vec{i}(-t)+\vec{j}(4t)=-2\vec{j}-3\vec{i}
    -t=-3\rightarrow t=3
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  2. #2
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    Re: Perpendicular Unit Vectors

    Your error is that you shouldn't set t \vec{v} + \vec{u} equal to the zero vector (in fact this has no solutions). All you need to do is set the \vec{i}-component to zero so that t\vec{v} + \vec{u} lies on the j-axis.

    We ignore the j-component, and obtain the equation

    t(-\vec{i}) + 3\vec{i} = 0

    t = 3
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  3. #3
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    Re: Perpendicular Unit Vectors

    Ok, thanks for clearing that up.
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