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Give the negative of the following vectors: A = 6.0 units SE and B = 7.5 units 67 0
N of W.

Sagot :

Onyebuchinnajiidn

The magnitude of the resultant of the two vectors when the position of vector A is known is [tex]R = \sqrt{92.25 - 90 cos(\theta)} \ \ \ units[/tex].

The given parameters;

  • vector A = 6 units SE
  • vector B = 7.5 units 67⁰ N W

The resultant of the two vectors is calculated by applying parallelogram law as shown below;

[tex]R^2 = a^2 + b^2 - 2ab \ cos(\theta)\\\\R^2 = (6)^2 + (7.5)^2 - 2(6 \times 7.5) cos(\theta)\\\\R^2 = 92.25 - 90 cos(\theta)\\\\R = \sqrt{92.25 - 90 cos(\theta)}[/tex]

Thus, the magnitude of the resultant of the two vectors when the position of vector A is known is [tex]R = \sqrt{92.25 - 90 cos(\theta)} \ \ \ units[/tex].

Learn more about parallelogram law here: https://brainly.com/question/23933745

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