draw each of the following vectors, label an angle that specifies the vectors direction, then find its magnitude and direction.a. B= -4.0 I+ 4.0jb. r= (-2.0i-1.0j) cmc. v= (-10-100j) m/s d. a= (20i+10j) m/s^2the I's and j's have the hat and its the vector simble on the letters
[tex]\begin{gathered} A) \\ B=-4.0i+4.0j \\ |B|=\sqrt{(-4.0)^2+(4.0)^2} \\ |B|=\sqrt[]{16^{}+16^{}} \\ |B|=\sqrt[]{32^{}} \\ |B|=4\sqrt{2} \\ \text{The magnitude of B is }4\sqrt[]{2} \\ \theta=\tan ^{-1}(\frac{4.0}{-4.0})=135\text{ \degree} \\ The\text{ angle of B is }135\text{ \degree} \\ B) \\ R=(-2.0i-1.0j)cm \\ |R|=\sqrt[]{(-2.0)^2+(-1.0)^2} \\ |R|=\sqrt{4.0+1.0} \\ |R|=\sqrt[]{5.0} \\ \text{The magnitude of R is }\sqrt[]{5.0}cm \\ \theta=\tan ^{-1}(\frac{-1.0}{-2.0})=26.57\text{ \degree, but it is below of negative x-axis, hence} \\ \theta=180+26.57=206.57\text{ \degree} \\ \text{The angle of R is }206.57\text{ \degree} \\ C) \\ V=(-10i-100j)\text{ m/s} \\ |V|=\sqrt{(-10)^2+(-100)^2} \\ |V|=\sqrt[]{100+10000} \\ |V|=\sqrt[]{10100}=10\sqrt{101}\approx100.5\text{ m/s} \\ \text{The magnitude of V is }100.5\text{ m/s} \\ \theta=\tan ^{-1}(\frac{-100}{-10})\approx264.29 \\ \text{The angle of V is }264.29\text{ \degree} \\ \\ D) \\ A=(20i+10j)m/s^2 \\ |A|=\sqrt{20^2+10^2} \\ |A|=\sqrt{400+100} \\ |A|=\sqrt{500}=10\sqrt{5}\approx22.36m/s^2 \\ \text{The magnitud of A is }22.36m/s^2 \\ \theta=\tan ^{-1}(\frac{10}{20})=26.57\text{ \degree} \\ \text{The angle of A is }26.57\text{ \degree} \end{gathered}[/tex]
