Respuesta :
By definition of vectors exist as quantities that contain magnitude (positive or negative) and direction.
A) The unit vector in the direction of [tex]$F = i + \frac{3j}{2} +2k.[/tex]
B) The distance between [tex]F_{1}[/tex] and [tex]F_{2}[/tex] exists 2.
C) i.) Required vector = −4j − 5k
ii.) Required vector = 5i − 6j − 8k
How to estimate unit vector?
To estimate the unit vector in the direction of V = 2i+3j+4k
|V| [tex]$= \sqrt{2^2+3^2+4^2}[/tex]
[tex]$= \sqrt{29}[/tex]
A) Unit vector
[tex]$F= \frac{V}{ |V|}[/tex]
[tex]$F= \frac{2i+3j +4k}{2}[/tex]
[tex]$F = \frac{2i}{2} + \frac{3j}{2} +\frac{4k}{2}[/tex]
[tex]$F= i + \frac{3j}{2} +2k.[/tex]
Therefore, unit vector in the direction of [tex]$F = i + \frac{3j}{2} +2k.[/tex]
B) To estimate the distance between [tex]F_{1}[/tex] = 3i+2j+ 5k and [tex]F_{2}[/tex] = 3i +4j +5k
[tex]$d = \sqrt{(3-3)^{2}+(4-2)^{2}+(5-5)^{2}}[/tex]
= 2
Therefore, the distance between [tex]F_{1}[/tex] and [tex]F_{2}[/tex] exists 2.
C) i.) P, (3,2,2), P₂(3,-2,-3)
Required vector =|[tex]P_{1}[/tex]P₂|=(3−3)i + ((−2)-2)j + ((-3)-2)k
= −4j − 5k
ii.) P₂ (2,-3,-6), [tex]P_{3}[/tex] (-3,3,2)
Required vector =|P₂[tex]P_{3}[/tex]|=(-3-2)i + (3-(-3))j + (2-(-6))k
= 5i − 6j − 8k
To learn more about vectors refer to:
https://brainly.com/question/2166498
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