# Perpendicular Unit Vectors

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

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

$\displaystyle t = 3$
• Jan 16th 2013, 01:37 PM
GrigOrig99
Re: Perpendicular Unit Vectors
Ok, thanks for clearing that up.