Solved Vector Worksheet Determine the numerical values of | Chegg.com - Free Printable

Let's solve each part of this Vector Worksheet step by step.

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Part 1: Determine the x- and y-components of each vector

We use:
- $ A_x = A \cos\theta $
- $ A_y = A \sin\theta $

But we must be careful about the angle and quadrant to get correct signs.

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#### Vector A: Magnitude = 5, Angle = 130° from positive x-axis
- This is in the second quadrant.
- $ A_x = 5 \cos(130^\circ) $
- $ A_y = 5 \sin(130^\circ) $

Using trigonometric values:
- $ \cos(130^\circ) = -\cos(50^\circ) \approx -0.6428 $
- $ \sin(130^\circ) = \sin(50^\circ) \approx 0.7660 $

So:
- $ A_x = 5 \times (-0.6428) = -3.214 $
- $ A_y = 5 \times 0.7660 = 3.830 $

✔ Answer:
- $ A_x = -3.21 $
- $ A_y = 3.83 $

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#### Vector B: Magnitude = 4, Angle = 30° below negative x-axis → 180° + 30° = 210° from positive x-axis? Wait!

Wait — angle shown is 30° below negative x-axis, so it’s in the third quadrant.

But the angle from positive x-axis is:
- $ 180^\circ + 30^\circ = 210^\circ $

Alternatively, you can think:
- From negative x-axis, going down 30° → angle with positive x-axis is $ 180^\circ + 30^\circ = 210^\circ $

So:
- $ B_x = 4 \cos(210^\circ) = 4 \cos(180^\circ + 30^\circ) = 4(-\cos(30^\circ)) = 4(-0.8660) = -3.464 $
- $ B_y = 4 \sin(210^\circ) = 4 \sin(180^\circ + 30^\circ) = 4(-\sin(30^\circ)) = 4(-0.5) = -2.0 $

✔ Answer:
- $ B_x = -3.46 $
- $ B_y = -2.00 $

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#### Vector C: Magnitude = 5, Angle = 45° below positive x-axis → fourth quadrant

So angle from positive x-axis is $ -45^\circ $ or $ 315^\circ $

- $ C_x = 5 \cos(-45^\circ) = 5 \cos(45^\circ) = 5 \times 0.7071 = 3.535 $
- $ C_y = 5 \sin(-45^\circ) = 5 \times (-\sin(45^\circ)) = 5 \times (-0.7071) = -3.535 $

✔ Answer:
- $ C_x = 3.54 $
- $ C_y = -3.54 $

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Part 2: Draw and label vectors from components; find magnitude and angle

Use:
- $ |\vec{V}| = \sqrt{V_x^2 + V_y^2} $
- $ \theta = \tan^{-1}(V_y / V_x) $, adjust for quadrant

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#### A: $ A_x = 3, A_y = -2 $
- This is in fourth quadrant
- $ A = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.606 $
- $ \theta = \tan^{-1}(-2/3) \approx -33.7^\circ $ → or $ 360^\circ - 33.7^\circ = 326.3^\circ $

✔ Answer:
- $ A = 3.61 $
- $ \theta = 326.3^\circ $ (or $ -33.7^\circ $)

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#### B: $ B_x = -2, B_y = 2 $
- Second quadrant
- $ B = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.828 $
- $ \theta = \tan^{-1}(2 / -2) = \tan^{-1}(-1) = -45^\circ $, but in second quadrant → $ 180^\circ - 45^\circ = 135^\circ $

✔ Answer:
- $ B = 2.83 $
- $ \theta = 135^\circ $

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#### C: $ C_x = 0, C_y = -2 $
- This is straight down on the negative y-axis
- $ C = \sqrt{0^2 + (-2)^2} = 2 $
- $ \theta = 270^\circ $ (or $ -90^\circ $)

✔ Answer:
- $ C = 2 $
- $ \theta = 270^\circ $

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Part 3: Components in different coordinate systems

#### First Coordinate System: Standard (x right, y up)

Vector $ \vec{A} $ has magnitude 5, at $ 30^\circ $ above x-axis.

So:
- $ A_x = 5 \cos(30^\circ) = 5 \times 0.8660 = 4.33 $
- $ A_y = 5 \sin(30^\circ) = 5 \times 0.5 = 2.5 $

✔ Answer:
- $ A_x = 4.33 $
- $ A_y = 2.50 $

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#### Second Coordinate System: Rotated (x-axis tilted)

The coordinate system is rotated such that:
- The new x-axis is along the dashed line (horizontal), and the new y-axis is vertical, but the whole system is rotated by $ 30^\circ $ clockwise?

Wait — look carefully:

In the second diagram:
- The dashed line is horizontal.
- The vector $ \vec{A} $ makes $ 30^\circ $ with the dashed line, which is now the new x-axis.
- The new y-axis is rotated $ 30^\circ $ from vertical.

So the vector $ \vec{A} $ is at $ 30^\circ $ above the new x-axis.

Therefore, in the rotated coordinate system, the components are:
- $ A_x' = 5 \cos(30^\circ) = 5 \times 0.8660 = 4.33 $
- $ A_y' = 5 \sin(30^\circ) = 5 \times 0.5 = 2.50 $

But wait — the question asks for $ A_x $ and $ A_y $ in the two coordinate systems shown.

Let’s clarify:

In the first system: standard Cartesian (x right, y up). We already did this:
→ $ A_x = 5 \cos(30^\circ) = 4.33 $, $ A_y = 5 \sin(30^\circ) = 2.50 $

In the second system, the axes are rotated 30° clockwise relative to the first one.

But the vector $ \vec{A} $ is still pointing at $ 30^\circ $ above the original x-axis.

Now, the new x-axis is rotated $ 30^\circ $ clockwise from the original x-axis.

So the angle between $ \vec{A} $ and the new x-axis is:
- Original direction: $ 30^\circ $ above x-axis
- New x-axis: $ -30^\circ $ from original x-axis
- So angle between vector and new x-axis: $ 30^\circ - (-30^\circ) = 60^\circ $

So in the rotated coordinate system, the vector makes $ 60^\circ $ with the new x-axis.

Thus:
- $ A_x' = 5 \cos(60^\circ) = 5 \times 0.5 = 2.5 $
- $ A_y' = 5 \sin(60^\circ) = 5 \times 0.8660 = 4.33 $

✔ Answer for rotated system:
- $ A_x = 2.50 $
- $ A_y = 4.33 $

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✔ Final Answers Summary

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#### Part 1: Components from magnitude and angle

| Vector | $ A_x $ | $ A_y $ |
|--------|----------|----------|
| A | -3.21 | 3.83 |
| B | -3.46 | -2.00 |
| C | 3.54 | -3.54 |

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#### Part 2: Magnitude and angle from components

| Vector | Magnitude | Angle $ \theta $ |
|--------|-----------|--------------------|
| A | 3.61 | 326.3° |
| B | 2.83 | 135° |
| C | 2.00 | 270° |

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#### Part 3: Components in two coordinate systems

System 1 (standard):
- $ A_x = 4.33 $
- $ A_y = 2.50 $

System 2 (rotated):
- $ A_x = 2.50 $
- $ A_y = 4.33 $

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🔍 Notes:

- Always check quadrants for sign.
- Use $ \cos $ and $ \sin $ based on angle from positive x-axis.
- For rotated systems, find the angle between vector and new axis.

Let me know if you'd like a sketch explanation!