Respuesta :
Answer:
Part (a) a + b = (8.0i - 2.0j) and [tex]\theta = 14.03^o[/tex] from the x-axis
Part (b) b - a = (2.0i - 6.0j) and [tex]\theta = -71.06^o[/tex] from the x- axis
Explanation:
Given,
- [tex]\vec{a}\ =\ (3.0m)i\ +\ (4.0m)j[/tex]
- [tex]\vec{b}\ =\ (5.0m)i\ +\ (-2.0m)j[/tex]
From the addition of the two vectors,
[tex]\vec{a}\ +\ \vec{b}\ =\ (3.0i\ +\ 4.0j)\ +\ (5.0i\ -\ 2.0j)\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ (3.0\ +\ 5.0)i\ +\(4.0\ -\ 2.0)j\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ 8.0i\ +\ 2.0j[/tex]
Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).
[tex]\therefore Tan\theta\ =\ \dfrac{2.0}{8.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{2.0}{8.0}\ \right )\\\Rightarrow \theta\ =\ 14.03^o[/tex]
Part (b)
From the subtraction of the two vectors,
[tex]\vec{b}\ -\ \vec{a}\ =\ (5.0i\ -\ 2.0j)\ -\ (3.0i\ +\ 4.0j)\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ (5.0\ +\ 3.0)i\ +\(-2.0\ -\ 4.0)j\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ 2.0i\ -\ 6.0j[/tex]
Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).
[tex]\therefore Tan\theta\ =\ \dfrac{-6.0}{2.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{-6.0}{2.0}\ \right )\\\Rightarrow \theta\ =\ -71.06^o[/tex]
