Question

022 A. Find the unit vector in the direction of F 2+3j+4k B. Find the distance between F1 = 3i+2j+ 5k and F2 = 3i +4j +5k C Find the component of each directed line segment whose initial point contains p, and the terminal point P₂ i.) P, (3,2,2), P₂(3,-2,-3) ii.) P₂(2,-3,-6), P3(-3,3,2) ​

Answer 1

By definition of vectors exist as quantities that contain magnitude (positive or negative) and direction.

A) The unit vector in the direction of $F = i + \frac{3j}{2} +2k.$

B) The distance between $F_{1}$ and $F_{2}$ 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| $= \sqrt{2^2+3^2+4^2}$

$= \sqrt{29}$

A) Unit vector

$F= \frac{V}{ |V|}$

$F= \frac{2i+3j +4k}{2}$

$F = \frac{2i}{2} + \frac{3j}{2} +\frac{4k}{2}$

$F= i + \frac{3j}{2} +2k.$

Therefore, unit vector in the direction of $F = i + \frac{3j}{2} +2k.$

B) To estimate the distance between $F_{1}$ = 3i+2j+ 5k and $F_{2}$ = 3i +4j +5k

$d = \sqrt{(3-3)^{2}+(4-2)^{2}+(5-5)^{2}}$

= 2

Therefore, the distance between $F_{1}$ and $F_{2}$ exists 2.

C) i.) P, (3,2,2), P₂(3,-2,-3)

Required vector =|$P_{1}$P₂|=(3−3)i + ((−2)-2)j ​+ ((-3)-2)k

= −4j − 5k

ii.) P₂ (2,-3,-6), $P_{3}$ (-3,3,2) ​

Required vector =|P₂$P_{3}$|=(-3-2)i + (3-(-3))j ​+ (2-(-6))k

= 5i − 6j − 8k

To learn more about vectors refer to:

brainly.com/question/2166498

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