Respuesta :
The magnitude of each vector is the Pythagorean sum of its components.
a. |v1| = √(2² + (-6)²) = √40 = 2√10
|v2| = √((-4)² + 7²) = √65
b. To make each vector into a unit vector, divide each component by the vector's magnitude.
u1 = v1/|v1| = (2/(2√10), -6/(2√10))
u1 = (√10/10, -3√10/10)
u2 = v2/|v2| = (-4/√65, 7/√65)
u2 = (-4√65/65, 7√65/65)
- a) [tex]|v_1| = \sqrt{40}, |v_2| = \sqrt{65}[/tex]
- b) [tex]v_{1u} = (\frac{2}{\sqrt{40}}, -\frac{6}{\sqrt{40}})[/tex]
- b) [tex]v_{2u} = (\frac{-4}{\sqrt{65}}, \frac{7}{\sqrt{65}})[/tex]
In item a, we find the magnitude of each vector, and then, we use this magnitude to find a unit vector in the same direction as the given vectors in item b.
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Magnitude of a vector:
The magnitude of a vector v = (x,y) is given by:
[tex]|v| = \sqrt{x^2 + y^2}[/tex]
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Item a:
v1=(2,-6), the magnitude of v1 is:
[tex]|v_1| = \sqrt{2^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40}[/tex]
v2=(-4,7), the magnitude of v2 is:
[tex]|v_2| = \sqrt{(-4)^2 + (7)^2} = \sqrt{16 + 49} = \sqrt{65}[/tex]
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Item b:
To find the unit vector, in the same direction, each component of the vector is divided by it's magnitude.
v1=(2,-6), magnitude [tex]\sqrt{40}[/tex]. Thus, the unit vectors in the direction of v1 is:
[tex]v_{1u} = (\frac{2}{\sqrt{40}}, -\frac{6}{\sqrt{40}})[/tex]
For v2, same thing:
[tex]v_{2u} = (\frac{-4}{\sqrt{65}}, \frac{7}{\sqrt{65}})[/tex]
A similar question is given at https://brainly.com/question/16250055
