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Unit Vectors and Direction AnglesIB Math HL Notes 6A.3Consider vector u=3i+4j. Find the magnitude u. If we divide the vector u by u = 5, we get another vector that is PARALLEL to u, since they are scalar multiples of each other. Find this new vector u5 = 35i+45jThis vector is a unit vector in the same direction. Prove that it is a unit vector.u5=352+452=9+1625=1Therefore, to find a unit vector in the same direction as a given vector, we divide that vector by its own magnitude.419745820266 This is tightly connected to the concept of the direction angle of a given vector. The direction angle of a vector (in standard position) is the angle it makes with the positive x-axis. Essentially, the x-component of a vector = ucosθAnd the y-component of a vector = usinθPractice 1: Find the direction angle of each vector, and find a vector of magnitude 7 that is parallel to each.(Hint: DRAW them)u=2i+2j b) v=3i-4ju=22v=52=22cosθ 3=5cosθ12=cosθ 35=cosθθ=π4 0.927 = θ7cosπ4i+7sinπ4j 7cos0.927i+7sin0.927j Or if we use unit vectors we get:7222i+7222 735i+7-45jPractice 2: Write each vector in component form with θ as the angle the vector makes with the positive horizontal axis.u=310, θ=62°?b) u=12, θ=135° (no calculator)u=145.536273.714u=12×2212×22=-6262Using vectors to model force, displacement and velocityExample: A plane is flying on a bearing of 170° at a speed of 840 km/h. The wind is blowing in the direction of N 120° E with a strength of 60 km/h. (Note: course is from N clockwise)Find the vector components of the plane’s still-air velocity and the wind’s velocity.Determine the true velocity (ground) of the plane in component form.Write down the true speed and direction of the plane. Let u be the plane’s still air velocity and v be the wind velocity, then:u=840cos280840sin280=145.864-827.239, v=60cos33060sin330=51.961-30 u+v=145.864-827.239+51.961-30=197.825-857.239Speed: u+v=197.8252+-857.2392=879.769 km/hDirection 197.825879.769=cosθ, θ=77.005 so the direction is a bearing of (77+90) 167 degrees.F//=Fgravsinθ Fperp=Fgravcosθ The inclined plane is a classic example of resultant forces. Use the “tilted head” approach to break the Force of gravity into its two components: The // component to the plane which causes the object to slide down, and the perpendicular component (when no friction is assumed, this is balanced by the support of the plane, i.e. the “normal” force).Thus, in essence we are only dealing with the // or horizontal tilted component.97790153670 θ θ Example: A box is being pulled up a 15° inclined plane. The force needed is 25 N. Find the horizontal and vertical components of the force vector and interpret each of them.484732537496Horizontal: 25NVertical: 96.6sin15=93.3 NExample: A 50 N weight is suspended by two strings as shown. Find the tensions T and S in the strings.48170332743Hint: make 2 equations by find the sums of the tensions for thehorizontal components = 50 cos90° and vertical components = 50 sin(-90°)scos135+tcos35=50cos90 or scos135+tcos35=0ssin135+tsin35=-50sin-90 or ssin135+tsin35=50Therefore:s=41.589, t=35.9015096358472891Example: A motor boat with the power to steer across a river at 30 km/h is moving such that the bow is pointed in a northerly direction. The stream is moving eastward at 6 km/h. The river is 1 km wide. Where on the opposite side will the boat meet the land? Since it would take 2 minutes to travel 1 km across it the current will move the boat 0.2 km east of the point directly across from the boat’s initial position.Suppose the boat wants to reach a point exactly north of the starting point. In which direction should the boat be steered in order to achieve this objective?sinθ=630θ=11.5°Example: Forces F = (-10, 3), G = (-4, 1) and H = (4, -10) act on a point P. Find the additional force required to keep the system in equilibrium.(10,6) ................
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