6.3b Direction Angles
If u is a unit vector such that θ is the angle (measured counterclockwise) from the positive x-axis to u, the terminal point of u lies on the unit circle and you have
u = < x,y > = < cos θ, sin θ > = (cos θ) i + (sin θ) j
Suppose that u is a unit vector with direction angle θ. If v is any vector that makes an angle θ with the positive x-axis, then it has the same direction as u and you can write:
v = llvll < cos θ, sin θ > = ll v ll (cos θ) i + ll v ll(sin θ) j
Because v = ai + bj = ll v ll(cos θ) i + ll v ll(sin θ) j
it follows that the direction angle θ for v is determined from
tan θ = (sin θ) ÷ (cos θ)
= (ll v llsin θ) ÷ (ll v ll cos θ)
= b÷ a
Example 1: Find the direction Angle of the vector u = 2i + 4j
The direction angle is tan θ = b ÷ a = 4 ÷ 2 = 2
so tan-1(2) = θ
therefore θ = 63.43494882° and since this vector is in the first quadrant it is the angle.
Example 2: Find the magnitude and direction angle of the vector v = 8 (cos 135°i + sin 135°j )
v = 8 cos 135°i + 8 sin 135°j
recall v = ai + bj = ll v ll(cos θ) i + llv ll(sin θ ) j
therefore ll v ll = 8 and θ = 135°
Example 3: find the component form of the sum of u and v with direction angle θ u and θv
ll u ll = 2 and θu = 30 ° and ll v ll = 2 and θ v = 90 °
Vector u = 2 cos 30 ° i + 2 sin 30 ° j
= 2 (√(3)/2) i + 2(1/2) j = √(3) i + j
u = < √3 , 1 >
Vector v = 2 cos 90 ° i + 2 sin 90 ° j
= 2(0) i + 2(1) j = 2 j
v = < 0 , 2 >
u + v = < √3 , 1 > + < 0 , 2 > = < √3 , 3 >
Use the Law of Cosines to find angle θ between the given vectors (0° ≤ θ ≤ 180° )
Example 4:
v = 3i + j and w = 2i - j
v = < 3 , 1 > and w = < 2 , -1 >
v + w = < 3 + 2 , 1 + -1 > = < 5 , 0 > so
the ll v + w ll = √(5 2 + 02) = 5
ll v ll = √( 3 2 + 12) = √(9 + 1) = √(10)
ll w ll = √( 2 2 + (-1)2) = √(4 + 1) = √(5)
ll v + w ll = 5
Using the Law of Cosines:
52 = (√(10)) 2 + (√(5))2 - 2 (√(10))(√(5))Cos θ
25 = 10 + 5 - 2 √(50) Cos θ
10 = - 2 √(50) Cos θ
-.7071067812 = Cos θ
135 ° = θ
Example 5: Find the angle between the forces given the magnitude of their resultant:
Force 1 = 3000 pounds
Force 2 = 1000 pounds
Resultant Force = 3750 pounds
37502 = 10002 + 30002 - 2(1000)(3000)Cos β
-.67708333333 = Cos β
β = 132.6161427 °
θ = 47.38385729°
