# Vector Addition as Net Effect

Suppose that a particle undergoes two displacements, the first one represented as \(\overrightarrow a \), which is 4 meters towards the east, and the second one represented as \(\overrightarrow b \), which is 4 meters towards the north:

To calculate the total distance travelled by the particle, we simply add the magnitudes of the two displacements: \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| = 8\) meters.

However, what is the *net effect* of the two displacements? The net effect is that relative to its initial position, the particle is at a distance of \(4\sqrt 2 \) meters in the north-east direction:

Thus, the *net displacement* of the particle is \(4\sqrt 2 \) meters, north-east. Observe that:

- the net distance travelled is a scalar quantity, and equal to \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| = 8\) meters.
- the net displacement is a vector quantity. Since it is the result of vectors \(\overrightarrow a \) and \(\overrightarrow b \) combined, we can represent the net displacement as \(\overrightarrow a + \overrightarrow b \). The value of this net displacement is \(4\sqrt 2 \) meters, north-east. Note that this + sign in \(\overrightarrow a + \overrightarrow b \)
*does not represent normal addition*, in the way we think about addition. It represents*vector addition*– it is calculating the*net effect*of two vectors.

To put it differently:

- . \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right|\). is your normal, scalar, addition. In this case, \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right|\, = \,8\) meters.
- \(\overrightarrow a + \overrightarrow b \) is vector addition, and is equal to the net effect of vectors \(\overrightarrow a \) and \(\overrightarrow b \). In this case, \(\overrightarrow a + \overrightarrow b \) \( = \,4\sqrt 2 \) meters, north-east.

**Exampl 1:** Consider three vectors:

\(\overrightarrow a \): 2 units, east

\(\overrightarrow b \): 2 units, north

\(\overrightarrow c \): 2 units, west

Find

(i) \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right|\) (ii) \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow c } \right|\) (iii) \(\left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|\)

(iv) \(\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|\) (v) \(\overrightarrow a + \overrightarrow b \)

(vi) \(\overrightarrow a + \overrightarrow c \) (vii) \(\overrightarrow b + \overrightarrow c \) (viii) \(\overrightarrow a + \overrightarrow b + \overrightarrow c \)

**Solution:** We note that \(\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 2\) units, and thus the answer for each of (i), (ii) and (iii) is 4 units, while the answer for (iv) is 6 units.

(v) Now, let us determine \(\overrightarrow a + \overrightarrow b \). Observe the following figure:

Clearly, \(\overrightarrow a + \overrightarrow b = 2\sqrt 2 \) units, north-east.

(vi) The following figure shows \(\overrightarrow a \) and \(\overrightarrow c \):

What will be the *net effect* of these two vectors? Intuitively, it should be obvious to you that they will *cancel each other out*, and thus the result will be a vector of magnitude 0 units. However, no particular direction will exist for this vector. We can write this as follows: \(\overrightarrow a + \overrightarrow c = 0\) units, unspecified direction, or \(\overrightarrow a + \overrightarrow c = \vec 0\)

(vii) The following figure shows \(\overrightarrow b \) and \(\overrightarrow c \), and the net effect of \(\overrightarrow b \) and \(\overrightarrow c \):

Clearly, \(\overrightarrow b + \overrightarrow c = 2\sqrt 2 \) units, north-west.

(viii) The following figure shows \(\overrightarrow a \), \(\overrightarrow b \) and \(\overrightarrow c \):

What is the net effect of all the three vectors taken together? Since \(\overrightarrow a \) and \(\overrightarrow c \) cancel the effect of each other out, we have:

\(\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow b = \,2\) units, north