Saturday, March 24, 2007

Precalculus 6.3b Direction Angles

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°