Vector Addition Practice Worksheet - Free Printable

Let's solve this Vector Components and Vector Addition Worksheet step by step.

We are given 8 vectors (A through H), each with magnitude and direction (angle measured from the positive x-axis, counterclockwise). The task is:

1. Find the x- and y-components of each vector.
2. Choose any two vectors and find their vector sum (resultant).

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🔷 Step 1: Vector Components

For any vector $ \vec{V} $ with magnitude $ V $ and angle $ \theta $ measured from the positive x-axis, the components are:

$$
V_x = V \cos(\theta) \\
V_y = V \sin(\theta)
$$

We’ll compute these for all vectors.

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#### ✔ Vector A: 15 m at 28°
$$
A_x = 15 \cos(28^\circ) \approx 15 \times 0.8829 = 13.24\, \text{m} \\
A_y = 15 \sin(28^\circ) \approx 15 \times 0.4695 = 7.04\, \text{m}
$$

✔ $ \vec{A} = (13.24, 7.04) $

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#### ✔ Vector B: 20 m at 30°
$$
B_x = 20 \cos(30^\circ) \approx 20 \times 0.8660 = 17.32\, \text{m} \\
B_y = 20 \sin(30^\circ) = 20 \times 0.5 = 10.00\, \text{m}
$$

✔ $ \vec{B} = (17.32, 10.00) $

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#### ✔ Vector C: 18 m at 124°
Note: 124° is in the second quadrant (between 90° and 180°)

$$
C_x = 18 \cos(124^\circ) \approx 18 \times (-0.4384) = -7.89\, \text{m} \\
C_y = 18 \sin(124^\circ) \approx 18 \times 0.8988 = 16.18\, \text{m}
$$

✔ $ \vec{C} = (-7.89, 16.18) $

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#### ✔ Vector D: 25 m at 39° above negative x-axis → but angle from positive x-axis?

Wait: The diagram shows 39° above the negative x-axis, so that’s:

$$
\theta = 180^\circ - 39^\circ = 141^\circ
$$

So:
$$
D_x = 25 \cos(141^\circ) \approx 25 \times (-0.7771) = -19.43\, \text{m} \\
D_y = 25 \sin(141^\circ) \approx 25 \times 0.6293 = 15.73\, \text{m}
$$

✔ $ \vec{D} = (-19.43, 15.73) $

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#### ✔ Vector E: 30 m at 224°
224° is in the third quadrant

$$
E_x = 30 \cos(224^\circ) \approx 30 \times (-0.6947) = -20.84\, \text{m} \\
E_y = 30 \sin(224^\circ) \approx 30 \times (-0.7193) = -21.58\, \text{m}
$$

✔ $ \vec{E} = (-20.84, -21.58) $

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#### ✔ Vector F: 23 m at 25° below negative x-axis?

Looking at the diagram: it's pointing down to the right, and angle is measured from vertical axis? Wait — let's clarify.

The angle is shown as 25° from the negative y-axis or from the negative x-axis?

But the diagram shows the angle between the vector and the vertical axis downward. But standard convention is from the positive x-axis.

From the diagram: vector F is in the fourth quadrant, and the angle is 25° from the negative y-axis toward the negative x-axis? No — wait.

Actually, looking at the diagram: the vector points downward to the right, and the angle is drawn from the negative y-axis to the vector, which means it's 25° from the negative y-axis toward the positive x-axis.

So total angle from positive x-axis:

- From positive x-axis to negative y-axis is 270°
- But better: think of it as being 90° + 25° = 115°? No.

Wait: if it's pointing down and to the right, and the angle is 25° from the vertical (y-axis), then:

- Angle from positive x-axis = 90° + 25° = 115°? No — that would be up and left.

Wait: if it's below the negative y-axis? No.

Actually, the vector is pointing down and to the right, so it's in the fourth quadrant.

If the angle is 25° from the vertical (negative y-axis) toward the positive x-axis, then:

- That means it makes a 25° angle with the negative y-axis, so its direction from the positive x-axis is:

$$
\theta = 270^\circ + 25^\circ = 295^\circ \quad \text{(no)} \\
\text{Wait: } 270^\circ \text{ is negative y-axis. If you go 25° toward positive x-axis, that's } 270^\circ - 25^\circ = 245^\circ?
$$

No — let's think carefully.

If the vector is pointing down and to the right, and the angle is 25° from the negative y-axis toward the positive x-axis, then:

- The angle from the positive x-axis is: $ 360^\circ - (90^\circ - 25^\circ) $? No.

Better: from positive x-axis, going clockwise or counterclockwise?

Standard: angles increase counterclockwise from positive x-axis.

So, if the vector is in the fourth quadrant, and makes 25° with the negative y-axis, then:

- The angle from the positive x-axis is: $ 270^\circ + 25^\circ = 295^\circ $? No — that’s in the fourth quadrant.

Wait: negative y-axis is 270°. Moving toward positive x-axis (i.e., to the right) from negative y-axis is clockwise, but we measure counterclockwise.

So actually, the vector is at an angle of:

$$
\theta = 360^\circ - 25^\circ = 335^\circ
$$

Because from positive x-axis, go counterclockwise to 335°, which is 25° above the negative y-axis? No — 335° is 25° below the positive x-axis.

Wait: 335° is in the fourth quadrant, 25° above the negative x-axis? No.

Let’s do this:

- 360° = positive x-axis
- 335° = 25° below positive x-axis → yes, that’s down and to the right.

But the diagram shows the angle from the vertical (y-axis), not horizontal.

Wait — the angle is labeled as 25° between the vector and the vertical line.

So: vector F is pointing down and to the right, and the angle between it and the negative y-axis is 25°.

So, the angle from the positive x-axis is:

- Start from positive x-axis: 0°
- Go clockwise to negative y-axis: 270°
- But since it's 25° from negative y-axis toward positive x-axis, that’s 25° from 270° toward 0° → so θ = 270° + 25° = 295°? No — that’s 25° above negative y-axis.

Wait: no.

If the vector is in the fourth quadrant, and makes 25° with the negative y-axis, then:

- The angle from the positive x-axis is: $ 360^\circ - (90^\circ - 25^\circ) $? Let's use trigonometry.

Alternatively: from the positive x-axis, the vector makes an angle of:

$$
\theta = 360^\circ - 25^\circ = 335^\circ
$$

Wait — if it’s 25° from the negative y-axis, and negative y-axis is 270°, then:

- Moving toward positive x-axis (to the right), so angle is $ 270^\circ + 25^\circ = 295^\circ $? But that’s in the third quadrant.

Wait: no.

Let me sketch mentally:

- Negative y-axis: straight down → 270°
- Vector goes down and to the right, so it's to the right of the negative y-axis
- So angle from positive x-axis: start from positive x-axis, go counterclockwise: pass 90° (up), 180° (left), 270° (down), then 270° + 25° = 295° → that’s 25° to the right of down → yes, that’s correct.

But 295° is in the fourth quadrant, because 270° to 360° is fourth quadrant.

Yes! So:

$$
\theta = 270^\circ + 25^\circ = 295^\circ
$$

So:

$$
F_x = 23 \cos(295^\circ) = 23 \cos(-65^\circ) = 23 \cos(65^\circ) \approx 23 \times 0.4226 = 9.72\, \text{m} \\
F_y = 23 \sin(295^\circ) = 23 \sin(-65^\circ) = -23 \sin(65^\circ) \approx -23 \times 0.9063 = -20.84\, \text{m}
$$

✔ $ \vec{F} = (9.72, -20.84) $

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#### ✔ Vector G: 35 m at 307°

307° is in the fourth quadrant

$$
G_x = 35 \cos(307^\circ) = 35 \cos(-53^\circ) = 35 \cos(53^\circ) \approx 35 \times 0.6018 = 21.06\, \text{m} \\
G_y = 35 \sin(307^\circ) = 35 \sin(-53^\circ) = -35 \sin(53^\circ) \approx -35 \times 0.7986 = -27.95\, \text{m}
$$

✔ $ \vec{G} = (21.06, -27.95) $

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#### ✔ Vector H: 10 m at 59° below negative x-axis?

Diagram shows: angle from negative x-axis down to vector.

So: negative x-axis is 180°, and vector is 59° below it → so total angle from positive x-axis:

$$
\theta = 180^\circ + 59^\circ = 239^\circ
$$

So:

$$
H_x = 10 \cos(239^\circ) = 10 \cos(180^\circ + 59^\circ) = -10 \cos(59^\circ) \approx -10 \times 0.3584 = -3.58\, \text{m} \\
H_y = 10 \sin(239^\circ) = 10 \sin(180^\circ + 59^\circ) = -10 \sin(59^\circ) \approx -10 \times 0.8572 = -8.57\, \text{m}
$$

✔ $ \vec{H} = (-3.58, -8.57) $

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📌 Summary of All Components

| Vector | Magnitude | Angle | $ V_x $ | $ V_y $ |
|--------|-----------|-------|----------|----------|
| A | 15 m | 28° | 13.24 | 7.04 |
| B | 20 m | 30° | 17.32 | 10.00 |
| C | 18 m | 124° | -7.89 | 16.18 |
| D | 25 m | 141° | -19.43 | 15.73 |
| E | 30 m | 224° | -20.84 | -21.58 |
| F | 23 m | 295° | 9.72 | -20.84 |
| G | 35 m | 307° | 21.06 | -27.95 |
| H | 10 m | 239° | -3.58 | -8.57 |

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🔷 Step 2: Add Any Two Vectors

Let’s pick A and B for simplicity.

$$
\vec{A} = (13.24, 7.04) \\
\vec{B} = (17.32, 10.00)
$$

Add components:

$$
R_x = 13.24 + 17.32 = 30.56\, \text{m} \\
R_y = 7.04 + 10.00 = 17.04\, \text{m}
$$

Now find magnitude and direction of resultant:

$$
R = \sqrt{R_x^2 + R_y^2} = \sqrt{30.56^2 + 17.04^2} \approx \sqrt{933.8 + 290.4} = \sqrt{1224.2} \approx 34.99\, \text{m} \approx 35.0\, \text{m}
$$

Direction:

$$
\theta = \tan^{-1}\left(\frac{17.04}{30.56}\right) \approx \tan^{-1}(0.5576) \approx 29.2^\circ
$$

✔ Resultant: $ \vec{R} = (30.56, 17.04) $, magnitude ≈ 35.0 m at 29.2°

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✔ Final Answer

#### Vector Components:

- A: $ (13.24, 7.04) $
- B: $ (17.32, 10.00) $
- C: $ (-7.89, 16.18) $
- D: $ (-19.43, 15.73) $
- E: $ (-20.84, -21.58) $
- F: $ (9.72, -20.84) $
- G: $ (21.06, -27.95) $
- H: $ (-3.58, -8.57) $

#### Example Vector Sum: $ \vec{A} + \vec{B} $

- Resultant: $ (30.56, 17.04) $
- Magnitude: 35.0 m
- Direction: 29.2° from positive x-axis

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You can choose any other pair (e.g., C + D, or G + H) for extra practice.

Let me know if you'd like the answers written neatly in a table format or need help with another pair!