Find the direction of u=-3i+8j (round to the nearest tenth)

[tex]\bf u=-3i+8j\implies u=\ \textless \ \stackrel{a}{-3}~~,~~\stackrel{b}{8}\ \textgreater \ \qquad \theta =tan^{-1}\left( \frac{b}{a} \right) \\\\\\ \theta =tan^{-1}\left( \frac{8}{-3} \right)\implies \theta \approx -69.4^o[/tex]

keeping in mind that the range of the inverse tangent function is from π/2 to -π/2, that gives us an angle in the IV Quadrant, however, notice "a" and "b", "a" is negative and "b" is positive, that means the II Quadrant.

so we're looking for a reference angle of 69.4° on the II Quadrant, and that'd be 180° - 69.4°, or 110.6°.

abidemiokin abidemiokin · Answer 2 · 2022-04-14T11:22:28+02:00

A vector quantity is known to have magnitude and direction. The direction of u=-3i+8j (round to the nearest tenth) is 110.6 degrees

How to calculate the direction of a vector?

The direction a vector is determined by taking the inverse of the ratio of the y axis to that of the x-axis as shown:

Ф = arctan(y/x)

Given

u=-3i+8j

y = 8 and x = -3

Substitute

Ф = arctan(-8/3)
Ф = arctan(2.6667)
Ф = -69.44

Since ethe angle is negative in the second quadrant , hence:

Ф = 180 - 69.44
Ф = 110.6 degrees

Hence the direction of u=-3i+8j (round to the nearest tenth) is 110.6 degrees

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