Direction Angle of a Vector

[nycmathguy]

Section 6.3

 

[MathLover1]

62.

||v||=sqrt(x^2+y^2)=sqrt((-5)^2+4^2)=sqrt(41) unit

Now it is clear that the head of the vector lies in the second quadrant, and we can use a little trigonometry to find out the angle.

tan (θ)=4/5

θ=90°+tan ^-1(4/5)≈2.46685rad≈141.34°

64. direction angle is correct
need magnitude

 

[nycmathguy]

62.

||v||=sqrt(x^2+y^2)=sqrt((-5)^2+4^2)=sqrt(41) unit

Now it is clear that the head of the vector lies in the second quadrant, and we can use a little trigonometry to find out the angle.

tan (θ)=4/5

θ=90°+tan ^-1(4/5)≈2.46685rad≈141.34°

64. direction angle is correct
need magnitude