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Thread: 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 $\displaystyle \vec{j}(4t)=-2\vec{j}\rightarrow t=-2$ because the number can only be 0 where $\displaystyle t\vec{v}+\vec{u}$ lies on the $\displaystyle \vec{j}$-axis?

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

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

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

    We ignore the j-component, and obtain the equation

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

    $\displaystyle t = 3$
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  3. #3
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    Re: Perpendicular Unit Vectors

    Ok, thanks for clearing that up.
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