Question

Given: A = (4, 0) B = (-6, 10) Unit vector in the direction of AB

Answer 1

`\bf \begin{cases} A=(4,0)\\ B=(-6,10) \end{cases}\qquad B-A\implies (-6-4~~,~~10-0)\implies \stackrel{\textit{component form}}{(-10,10)=AB} \\\\[-0.35em] ~\dotfill\\\\ ||AB||\implies √((-10)^2+(10)^2)\implies √(100+100)\implies √(2(100)) \\\\\\ √(2(10^2))\implies 10√(2) \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{unit vector}}{\cfrac{AB}AB}\implies \cfrac{(-10,10)}{10√(2)}\implies \left( \cfrac{\stackrel{-1}{~~\begin{matrix} -10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~√(2)}~,~\cfrac{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~√(2)} \right) \implies \left(\cfrac{-1}{√(2)}~,~\cfrac{1}{√(2)} \right)`

`\bf \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{and rationalizing the denominator}}{\left( -\cfrac{√(2)}{2}~,~\cfrac{√(2)}{2} \right)}~\hfill`