Respuesta :
1) (-5, 5), 2) (7, -5), 3) (- 3, 28), 4) (- 12, 0), 5) 10, 6) 14, 7) (- 9, 9), 8) [tex]\sqrt{26}[/tex], 9) - 2, 10) 10, 11) [tex]\left(\frac{\sqrt{26}}{26}, \frac{5\sqrt{26}}{26}\right)[/tex], 12) [tex]\left(\frac{21\sqrt{10}}{10}, -\frac{7\sqrt{10}}{10} \right)[/tex], 13) (0, 7), 14) [tex](- 3\sqrt{3}, - 3)[/tex], 15) The vectors [tex]\vec a[/tex] and [tex]\vec b[/tex] are orthogonal, 16) [tex]7\sqrt{2}[/tex], [tex]135^{\circ}[/tex], 17) [tex]17[/tex], [tex]137^{\circ}[/tex], 18) [tex]12[/tex], [tex]240^{\circ}[/tex], 19) [tex](19.700, 15.392)[/tex]
How to analyze vectors
Vectors are elements that are charaterized for having a magnitude and a direction, whereas scalar are only characterized by magnitude. In this question we must apply different vector properties to obtain the corresponding results.
The first two exercises consist in calculating each vector based on its initial and terminal points, whose results are obtained below by sum of vectors with initial points at origin:
1) (-1, 3) - (4, -2) = (-5, 5)
2) (2, -6) - (- 5, -1) = (7, -5)
The following eight exercises are resolved by applying the operations of addition of vectors and multiplication of a vector by a scalar, and the definition of norm and dot product as well:
3) 3 · (3, 4) + 2 · (- 6, 8) = (9, 12) + (- 12, 16) = (- 3, 28)
4) (- 6, 8) - 2 · (3, 4) = (- 12, 0)
5) [tex]\| \vec v\| = \sqrt{(-6)^{2}+8^{2}}[/tex] = 10
6) [tex]\vec u \,\bullet \,\vec v =[/tex] 3 · (- 6) + 4 · 8 = 14
7) 3 · [(0, 2) - (3, -1)] = 3 · (- 3, 3) = (- 9, 9)
8) [tex]\|\vec v\| = \sqrt{1^{2}+5^{2}} = \sqrt{26}[/tex]
9) [tex]\vec u \,\bullet \,\vec v =[/tex] 3 · 1 + (- 1) · 5 = - 2
10) [tex]\vec w \,\bullet \, \vec v =[/tex] 0 · 1 + 5 · 2 = 10
11) The unit vector is obtained by dividing the vector by its magnitude:
[tex]\vec u_{v} = \frac{1}{\sqrt{26}}\cdot \left(1, 5\right) = \left(\frac{1}{\sqrt{26}}, \frac{5}{\sqrt{26}} \right) = \left(\frac{\sqrt{26}}{26}, \frac{5\sqrt{26}}{26}\right)[/tex]
12) The vector is equal to the product of the magnitude and its unit vector:
[tex]\vec r = 7 \cdot \frac{\vec u}{\|\vec u\|}[/tex]
[tex]\vec r = \frac{7}{\sqrt{10}}\cdot (3, -1) = \left(\frac{21}{\sqrt{10}}, -\frac{7}{\sqrt{10}} \right) = \left(\frac{21\sqrt{10}}{10}, -\frac{7\sqrt{10}}{10} \right)[/tex]
13) and 14) The vector is the product of the magnitude and the directional vector:
13) [tex]\vec r = 7 \cdot (\cos 90^{\circ}, \sin 90^{\circ}) = (0, 7)[/tex]
14) [tex]\vec r = 6\cdot (\cos 210^{\circ}, \sin 210^{\circ}) = (- 3\sqrt{3}, - 3)[/tex]
15) Two vectors are orthogonal if and only if their dot product is zero:
[tex](3.2, 4.8) \,\bullet\,(-3, 2) =[/tex] 3.2 · (- 3) + 4.8 · 2 = 0
The vectors [tex]\vec a[/tex] and [tex]\vec b[/tex] are orthogonal.
The next three exercises are resolved by definitions of norm and direction:
16) [tex]\|\vec v\| = \sqrt{(-7)^{2}+7^{2}} = 7\sqrt{2}[/tex], [tex]\theta = 135^{\circ}[/tex]
17) [tex]\|\vec v\| = 17[/tex], [tex]\theta = 137^{\circ}[/tex]
18) [tex]\|\vec q\| = 12[/tex], [tex]\theta = 240^{\circ}[/tex]
The last exercise consist in defining the vector as a product of magnitude and direction vector:
19) [tex]\vec v = 25\cdot (\cos 38^{\circ}, \sin 38^{\circ}) = (19.700, 15.392)[/tex]
To learn more on vectors, we kindly invite to check this verified question: https://brainly.com/question/13322477
