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$g: \begin{cases} x=3s \\ y=0 \\ z=1-4s \end{cases}$
How do I find the coordinates of a vector $e_2$ which is parallel to g, and $|e_2|=1?$
I know that $\sqrt{x_1^2+x_2^2+x_3^2}=1$.
I assume that the line is passing through a point (0,0,1) and it has direction the vector (3,0,-4).
So is it alright just to use the second coordinates (the ones in front of the scalar s) and make sure the norm is 1?
$e_2(\frac{3}{5}, 0, -\frac{4}{5})$
Is this an ok solution?
1 Answer
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HINT: the straight line is given by $$[x,y,z]=[0,0,1]+s[3,0,-4]$$ thus our serached vector is $$[3m,0,-4m]$$ with $\sqrt{9m^2+16m^2}=1$ from here you will get $m$
