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3.4: Coordinate Systems and Components of a Vector (Part 1)

  • Page ID
    17866
  • Vector A is shown in the x y coordinate system and extends from point b at A’s tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b.
    Figure 2.18. The quotient of the adjacent side Ax to the hypotenuse A is the cosine function of direction angle \(\theta_{A}\), Ax/A = cos \(\theta_{A}\), and the quotient of the opposite side Ay to the hypotenuse A is the sine function of \(\theta_{A}\), Ay/A = sin \(\theta_{A}\). When magnitude A and direction \(\theta_{A}\) are known, we can solve these relations for the scalar components:

    $$\begin{cases} A_{x} = A \cos \theta_{A} \\ A_{y} = A \sin \theta_{A} \ldotp \end{cases} \label{2.17}$$

    When calculating vector components with Equation 2.17, care must be taken with the angle. The direction angle \(\theta\)A of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwise measurement gives a negative angle.

    Example \(\PageIndex{3}\): Components of Displacement Vectors

    A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction 30° west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns 23° west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacement vectors in vector component form for each leg.

    Strategy

    Let’s adopt a rectangular coordinate system with the positive x-axis in the direction of geographic east, with the positive y-direction pointed to geographic north. Explicitly, the unit vector \(\hat{i}\) of the x-axis points east and the unit vector \(\hat{j}\) of the y-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use Equation 2.17 to find the scalar components of the displacements and Equation 2.12 for the displacement vectors.

    Solution

    On the first leg, the displacement magnitude is L1 = 200.0 m and the direction is southeast. For direction angle \(\theta_{1}\) we can take either 45° measured clockwise from the east direction or 45° + 270° measured counterclockwise from the east direction. With the first choice, \(\theta_{1}\) = −45°. With the second choice, \(\theta_{1}\) = + 315°. We can use either one of these two angles. The components are

    $$ L_{1x} = L_{1} \cos \theta_{1} = (200.0\; m) \cos 315^{o} = 141.4\; m,$$

    $$ L_{1y} = L_{1} \sin\theta_{1} = (200.0\; m) \sin 315^{o} = -141.4\; m,$$

    The displacement vector of the first leg is

    $$\vec{L}_{1} = L_{1x}\; \hat{i} + L_{1y}\; \hat{j} = (14.4\; \hat{i} - 141.4\; \hat{j})\; m \ldotp$$

    On the second leg of Trooper’s wanderings, the magnitude of the displacement is L2 = 300.0 m and the direction is north. The direction angle is \(\theta_{2}\) = + 90°. We obtain the following results:

    $$ L_{2x} = L_{2} \cos \theta_{2} = (300.0\; m) \cos 90^{o} = 0.0,$$

    $$ L_{2y} = L_{2} \sin \theta_{2} = (300.0\; m) \sin 90^{o} = 300.0\; m,$$

    $$\vec{L}_{2} = L_{2x}\; \hat{i} + L_{2y}\; \hat{j} = (300.0\; m)\; \hat{j} \ldotp$$

    On the third leg, the displacement magnitude is L3 = 50.0 m and the direction is 30° west of north. The direction angle measured counterclockwise from the eastern direction is \(\theta\)3 = 30° + 90° = + 120°. This gives the following answers:

    $$ L_{3x} = L_{3} \cos \theta_{3} = (50.0\; m) \cos 120^{o} = -25.0\; m,$$

    $$ L_{3y} = L_{3} \sin \theta_{3} = (50.0\; m) \sin 120^{o} = + 43.3\; m,$$

    $$\vec{L}_{3} = L_{3x}\; \hat{i} + L_{3y}\; \hat{j} = (-25.0\; \hat{i} + 43.3\; \hat{j})\; m \ldotp$$

    On the fourth leg of the excursion, the displacement magnitude is L4 = 80.0 m and the direction is south. The direction angle can be taken as either \(\theta_{4}\) = −90° or \(\theta_{4} = + 270°. We obtain

    $$ L_{4x} = L_{4} \cos \theta_{4} = (80.0\; m) \cos (-90^{o}) = 0,$$

    $$ L_{4y} = L_{4} \sin \theta_{4} = (80.0\; m) \sin (-90^{o}) = -80.0\; m,$$

    $$\vec{L}_{4} = L_{4x}\; \hat{i} + L_{4y}\; \hat{j} = (-80.0\; m)\; \hat{j} \ldotp$$

    On the last leg, the magnitude is L5 = 150.0 m and the angle is \(\theta_{5}\) = −23° + 270° = + 247° (23° west of south), which gives

    $$ L_{5x} = L_{5} \cos \theta_{5} = (150.0\; m) \cos 247^{o} = -58.6\; m,$$

    $$ L_{5y} = L_{5} \sin \theta_{5} = (150.0\; m) \sin 247^{o} = -138.1\; m,$$

    $$\vec{L}_{5} = L_{5x}\; \hat{i} + L_{5y}\; \hat{j} = (-58.6\; \hat{i} - 138.1\; \hat{j})\; m \ldotp$$

    Exercise 2.6

    If Trooper runs 20 m west before taking a rest, what is his displacement vector?

    Contributors

    • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).