Question

Determine whether u and v are equivalent. Explain.

Answer 1

hmmm the magnitude of a vector is simply its length, if we have the start and end points for it, we can use that to get it, is pretty much the same as getting the distance from one to the other, so let's get their distance for each.

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{5})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ ||u||=√((~~5 - 2~~)^2 + (~~5 - 4~~)^2)\implies ||u||=√( (3)^2 + (1)^2) \\\\\\ ||u||=√( 9 + 1)\implies ||u||=√( 10 ) \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ ||v||=√((~~3 - 0~~)^2 + (~~1 - 0~~)^2)\implies ||v||=√( (3)^2 + (1)^2) \\\\\\ ||v||=√( 9 + 1)\implies ||v||=√( 10 ) \\\\[-0.35em] ~\dotfill`

`\stackrel{ \textit{\LARGE u} }{(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{5})} ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{5}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{2}}} \implies \cfrac{ 1 }{ 3 } \\\\[-0.35em] ~\dotfill`

`\stackrel{ \textit{\LARGE v} }{(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1})} ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{1}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{0}}} \implies \cfrac{ 1 }{ 3 } \\\\\\ ~\hfill~\textit{\LARGE equivalent~~\checkmark}~\hfill~`