Adding vectors by components

With the components tool from §3.3, vector addition reduces to simple scalar addition per axis. More accurate than graphical and scalable to many vectors.

The golden rule

If \vec a = a_x \hat i + a_y \hat j and \vec b = b_x \hat i + b_y \hat j, then:

$$
\boxed{\vec a + \vec b = (a_x + b_x)\hat i + (a_y + b_y)\hat j}
$$

In 3D: add a z equation. Done.

Why: \hat i and \hat j are independent (orthogonal). So we add x and y components separately.

4-step strategy

For any vector-addition problem:

1. Choose axes — usually x horizontal, y vertical. For inclined-plane problems, a tilted axis often helps.
2. Decompose each vector — a_x = a \cos\theta_a, a_y = a \sin\theta_a, and similarly for \vec b.
3. Add corresponding components — c_x = a_x + b_x, c_y = a_y + b_y.
4. Convert back to magnitude and angle — c = \sqrt{c_x^2 + c_y^2}, \theta_c = \arctan(c_y / c_x).

Example 1 — three forces on an object

Scenario: an object subject to three forces:
– \vec F_1 = 30 N at 0° (east)
– \vec F_2 = 40 N at 90° (north)
– \vec F_3 = 20 N at 150° (northwest)

Net force?

Decompose:

$$
F_{1x} = 30 \cos 0° = 30,\quad F_{1y} = 30 \sin 0° = 0
$$

$$
F_{2x} = 40 \cos 90° = 0,\quad F_{2y} = 40 \sin 90° = 40
$$

$$
F_{3x} = 20 \cos 150° = -17.3,\quad F_{3y} = 20 \sin 150° = 10
$$

Add:

$$
F_x = 30 + 0 – 17.3 = 12.7\ \mathrm{N}
$$

$$
F_y = 0 + 40 + 10 = 50\ \mathrm{N}
$$

Magnitude:

$$
F = \sqrt{12.7^2 + 50^2} = \sqrt{161 + 2500} = \sqrt{2661} \approx 51.6\ \mathrm{N}
$$

Angle:

$$
\theta = \arctan(50 / 12.7) \approx 75.7°
$$

Net force ~51.6 N at ~75.7° (mostly north, slightly east).

Example 2 — recovering from components

Scenario: total displacement \vec d = -4\hat i + 3\hat j\ \mathrm{m}. Magnitude and angle?

$$
d = \sqrt{(-4)^2 + 3^2} = \sqrt{25} = 5\ \mathrm{m}
$$

$$
\tan\theta = 3 / (-4) = -0.75
$$

Calculator gives \theta \approx -36.9°. But components say d_x < 0, d_y > 0 — that’s quadrant II. So the correct angle:

$$
\theta = 180° – 36.9° = 143.1°
$$

Lesson: always cross-check the angle with the signs.

Subtraction

$$
\vec a – \vec b = (a_x – b_x)\hat i + (a_y – b_y)\hat j
$$

Same component rule — just flip the sign of \vec b.

Scalar multiplication

If k is a scalar, k \vec a gives another vector:

$$
k \vec a = (k a_x) \hat i + (k a_y) \hat j
$$

• k > 0: direction unchanged, magnitude scaled by |k|
• k < 0: direction reversed, magnitude scaled by |k|
• k = 0: zero vector

A few notes and common mistakes

1. Be precise with signs.
cos(120°) = -0.5, not +0.5. Check calculator mode (degrees vs radians).

2. For inclined-plane problems, use tilted axes.
If an object moves on a slope, take x along the slope and y perpendicular to it. Decompose gravity as mg\sin\theta and mg\cos\theta.

3. What does a zero component mean?
If \vec c = 0\hat i + 3\hat j, the vector points along y. If both are zero: zero vector.

4. 3D isn’t harder.
Adding a z component adds one equation — nothing more complex.

What you should be able to do

After this section, you should be able to:

• Write the component-addition rule and explain why it works
• Apply the 4-step strategy to 3–5-vector problems
• Recognize which quadrant gives the correct angle
• Compute scalar-vector multiplication and know its effect on direction and magnitude

Preview of §3.5

Now: how do we multiply two vectors? There are two kinds of vector multiplication:
– Scalar (dot) product: yields a scalar. Used for projection and work.
– Vector (cross) product: yields a vector. Used for torque and angular momentum.

§3.5 introduces the dot product.

📚 See also: Halliday Vol 1, Ch 3, §3.4 — Adding Vectors by Components.