Perpendicular Unit Vectors

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January 16th 2013, 10:38 AM
GrigOrig99

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: $t(-\vec{i}+4\vec{j})+3\vec{i}+2\vec{j}=0$
$-t\vec{i}+4t\vec{j}=-3\vec{i}-2\vec{j}$
$\vec{i}(-t)+\vec{j}(4t)=-2\vec{j}-3\vec{i}$
$-t=-3\rightarrow t=3$
January 16th 2013, 12:16 PM
richard1234

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

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

$t = 3$
January 16th 2013, 01:37 PM
GrigOrig99

Re: Perpendicular Unit Vectors

Ok, thanks for clearing that up.

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