1. A vector \vec xx is a unit vector if its magnitude is 1. Given \begin{gathered}\vec A = \dfrac{\sqrt3}2\,\vec\imath - \dfrac12\,\vec\jmath \vec B = -\dfrac{\sqrt2}2\,\vec\imath - \dfrac{\sqrt2}2\,\vec\jmath \vec C = \dfrac12\,\vec\imath - \dfrac{\sqrt3}2\,\vec\jmath - \dfrac12\,\vec k\end{gathered} \vec AA and \vec BB are unit vectors, while \vec CC is not, since \begin{gathered}\|\vec A\| = \sqrt{\left(\dfrac{\sqrt3}2\right)^ 2 + \left(-\dfrac12\right)^2} = 1 \|\vec B\| = \sqrt{\left(-\dfrac{\sqrt2}2\right) ^2 + \left(-\dfrac{\sqrt2}2\right)^2} = 1 \|\vec C\| = \sqrt{\left(\dfrac12\right)^2+\left (-\dfrac{\sqrt3}2\right)^2+\left(-\ dfrac12\right)^2} =\dfrac{\sqrt5}2 \neq 1\end{gathered}∥A∥=(23)2+(−21)2=1∥B ∥=(−22)2+(−22)2=1∥C∥=(21)2+(−23)2+( −21)2=25=1 2. Given some vector \vec xx , you can get the unit vector in the same direction as If \begin{gathered}\vec A = -12\,\vec\imath + 9\,\vec\jmath \vec B = 15\,\vec\imath-20\,\vec\jmath\end{g athered} then the unit vectors in the direction of \vec AA and \vec BB , respectively, are \begin{gathered}\dfrac{\vec A}{\|\vec A\|} = \dfrac{-12\,\vec\imath+9\,\vec\jmat h}{\sqrt{(-12)^2+9^2}} = \dfrac{-12\,\vec\imath+9\,\vec\jmat h}{15} = -\dfrac45\,\vec\imath + \dfrac35\,\vec\jmath \dfrac{\vec B}{\|\vec B\|} = \dfrac{15\,\vec\imath-20\,\vec\jmat h}{\sqrt{15^2+(-20)^2}} = \dfrac{15\,\vec\imath-20\,\vec\jmat h}{25} = \dfrac35\,\vec\imath - \dfrac45\,\vec\jmath\end{gathered} 3. \begin{gathered}\vec A = \|\vec A\| \left(\cos(180^\circ-\theta)\,\vec\ imath + \sin(180^\circ-\theta)\,\vec\jmath\ right) \vec A = 25 \left(-\sin(37^\circ)\,\vec\imath + \cos(37^\circ)\,\vec\jmath\right) \vec A = -15\,\vec\imath + 20\,\vec\jmath\end{gathered} \begin{gathered}\vec B = \|\vec B\| \left(\cos(180^\circ + \alpha)\,\vec\imath + \sin(180^\circ+\alpha)\,\vec\jmath\ right) \vec B = 75 \left(-\sin(53^\circ)\,\vec\imath-\ cos(53^\circ)\,\vec\jmath\right) \vec B = -60\,\vec\imath - 45\,\vec\jmath\end{gathered}