1. Unit vectors parallel

Hi there,

I need to find two unit vectors parallel to line L, where L is given by x = (z-1)/2, y = 1

Not sure how to approach the problem, could anybody help?

Thanks!

2. Re: Unit vectors parallel

Start by putting the line into its vector form. You need a parameter, say \displaystyle \begin{align*} t \end{align*}. So you can set \displaystyle \begin{align*} x = t \end{align*}, which means \displaystyle \begin{align*} \frac{z - 1}{2} = t \implies z = 2t + 1 \end{align*}.

So the line can be written as \displaystyle \begin{align*} L = ( t, 1, 2t + 1 ) = t (1 , 0 , 2 ) + ( 0, 1, 1 ) \end{align*}.

So the direction vector of the line is \displaystyle \begin{align*} (1, 0, 2) \end{align*}. What is the unit vector going in that direction? What is another unit vector that is parallel to this one?

3. Re: Unit vectors parallel

Originally Posted by Prove It
Start by putting the line into its vector form. You need a parameter, say \displaystyle \begin{align*} t \end{align*}. So you can set \displaystyle \begin{align*} x = t \end{align*}, which means \displaystyle \begin{align*} \frac{z - 1}{2} = t \implies z = 2t + 1 \end{align*}.

So the line can be written as \displaystyle \begin{align*} L = ( t, 1, 2t + 1 ) = t (1 , 0 , 2 ) + ( 0, 1, 1 ) \end{align*}.

So the direction vector of the line is \displaystyle \begin{align*} (1, 0, 2) \end{align*}. What is the unit vector going in that direction? What is another unit vector that is parallel to this one?
Ahh I get it now. So the unit vectors parallel would be +(1/sqrt5)(1, 0, 2) and -(1/sqrt5)(1, 0, 2)?

4. Re: Unit vectors parallel

They're fine, you could also move them around if you want...