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To find the component form of the vector \(\mathbf{v}\) with a magnitude of 4 and a bearing of 125°, we need to convert the polar coordinates (magnitude and direction) into Cartesian coordinates (components along the \(i\) and \(j\) directions). 1. **Convert the bearing to a standard angle:** Bearings are measured clockwise from the north direction. A bearing of 125° means the angle is 125° clockwise from the north. To convert this to a standard angle (measured counterclockwise from the positive x-axis), we use: \[ \text{Standard angle} = 360° - 125° = 235° \] 2. **Calculate the components:** The components of the vector can be found using trigonometry: \[ v_x = v \cos(\theta) \] \[ v_y = v \sin(\theta) \] where \(v\) is the magnitude and \(\theta\) is the standard angle. Given: \[ v = 4, \quad \theta = 235° \] Converting 235° to radians: \[ \theta = 235° \times \frac{\pi}{180°} = \frac{235\pi}{180} \approx 4.1015 \text{ radians} \] Now, calculate the components: \[ v_x = 4 \cos(235°) = 4 \cos(4.1015) \approx 4 \times (-0.5736) \approx -2.29 \] \[ v_y = 4 \sin(235°) = 4 \sin(4.1015) \approx 4 \times (-0.8192) \approx -3.28 \] 3. **Write the vector in component form:** \[ \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} = -2.29\mathbf{i} - 3.28\mathbf{j} \] Therefore, the correct answer is: \[ \boxed{E} \]

Question

To find the component form of the vector v\mathbf{v} with a magnitude of 4 and a bearing of 125°, we need to convert the polar coordinates (magnitude and direction) into Cartesian coordinates (components along the ii and jj directions). 1. Convert the bearing to a standard angle: Bearings are measured clockwise from the north direction. A bearing of 125° means the angle is 125° clockwise from the north. To convert this to a standard angle (measured counterclockwise from the positive x-axis), we use: Standard angle=360°125°=235° \text{Standard angle} = 360° - 125° = 235° 2. Calculate the components: The components of the vector can be found using trigonometry: vx=vcos(θ) v_x = v \cos(\theta) vy=vsin(θ) v_y = v \sin(\theta) where vv is the magnitude and θ\theta is the standard angle. Given: v=4,θ=235° v = 4, \quad \theta = 235° Converting 235° to radians: θ=235°×π180°=235π1804.1015 radians \theta = 235° \times \frac{\pi}{180°} = \frac{235\pi}{180} \approx 4.1015 \text{ radians} Now, calculate the components: vx=4cos(235°)=4cos(4.1015)4×(0.5736)2.29 v_x = 4 \cos(235°) = 4 \cos(4.1015) \approx 4 \times (-0.5736) \approx -2.29 vy=4sin(235°)=4sin(4.1015)4×(0.8192)3.28 v_y = 4 \sin(235°) = 4 \sin(4.1015) \approx 4 \times (-0.8192) \approx -3.28 3. Write the vector in component form: v=vxi+vyj=2.29i3.28j \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} = -2.29\mathbf{i} - 3.28\mathbf{j} Therefore, the correct answer is: E \boxed{E}

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Solution

To find the component form of the vector vv with a magnitude of 4 and a bearing of 125°, we need to convert the polar coordinates (magnitude and direction) into Cartesian coordinates (components along the ii and jj directions).

  1. Convert the bearing to a standard angle: Bearings are measured clockwise from the north direction. A bearing of 125° means the angle is 125° clockwise from the north. To convert this to a standard angle (measured counterclockwise from the positive x-axis), we use: Standard angle=360°125°=235° \text{Standard angle} = 360° - 125° = 235°

  2. Calculate the components: The components of the vector can be found using trigonometry: vx=vcos(θ) v_x = v \cos(\theta) vy=vsin(θ) v_y = v \sin(\theta) where vv is the magnitude and θ\theta is the standard angle. Given: v=4,θ=235° v = 4, \quad \theta = 235° Converting 235° to radians: θ=235°×π180°=235π1804.1015 radians \theta = 235° \times \frac{\pi}{180°} = \frac{235\pi}{180} \approx 4.1015 \text{ radians} Now, calculate the components: vx=4cos(235°)=4cos(4.1015)4×(0.5736)2.29 v_x = 4 \cos(235°) = 4 \cos(4.1015) \approx 4 \times (-0.5736) \approx -2.29 vy=4sin(235°)=4sin(4.1015)4×(0.8192)3.28 v_y = 4 \sin(235°) = 4 \sin(4.1015) \approx 4 \times (-0.8192) \approx -3.28

  3. Write the vector in component form: v=vxi+vyj=2.29i3.28j \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} = -2.29\mathbf{i} - 3.28\mathbf{j}

Therefore, the component form of the vector is 2.29i3.28j-2.29\mathbf{i} - 3.28\mathbf{j}.

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