Magnitude, Direction of a Vector and Unit Vector

Learn what vectors are, understand how they are represented, then explore vector operations — one step at a time.

Magnitude, Direction of a Vector and Unit Vector

A vector is a quantity that has both magnitude (size) and direction. If a vector is written in component form as

$$\overrightarrow{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$

then $a_1$, $a_2$, and $a_3$ are real numbers called the components of the vector, and $\hat{i}$, $\hat{j}$, $\hat{k}$ are the unit vectors along the $x$, $y$, and $z$ axes.

The same vector can be described in different ways depending on what information we want to show. The three important ideas are:

  1. Magnitude of a vector
  2. Direction of a vector
  3. Unit vector

Magnitude of a Vector

Step 1 of 7
1 square = 1 unit X Y O A(2, 1) B(4, 3) M(3, 0) OM = 3 2 2 AB = 2√2 A(x, y) y x OA = √(x² + y²)

Derivation Steps

The length of a directed line segment representing a vector is called the magnitude of that vector. The magnitude of vector a is written |a| and read as "the absolute value of vector a".

In the figure, A(2, 1), B(4, 3) and M(3, 0) are points, and O is the origin (0, 0). The lengths of OM, AB, OA and PQ are found using the distance formula.

Length of OM = √(3 – 0)² + (0 – 0)²

= √(3)² + (0)² = √9 + 0 = √9 = 3 unit

Length of AB = √(4 – 2)² + (3 – 1)²

= √(2)² + (2)² = √4 + 4 = √8 = 2√2

Therefore, the magnitude of vector AB is |AB| = 2√2 unit.

Similarly, for a point A(x, y), the length of OA = √(x – 0)² + (y – 0)²

= √(x)² + (y)² = √x² + y² unit

Therefore, the magnitude of vector OA is |OA| = √x² + y² unit.

Likewise, for P(x1, y1) and Q(x2, y2), the length of PQ = √(x2 – x1)² + (y2 – y1)² unit.

Therefore, |PQ| = √(x2 – x1)² + (y2 – y1)² unit.

In general, the magnitude of a vector is:

| a | = √(X-component)² + (Y-component)²

Since length is always expressed using a positive sign, the magnitude of a vector is also always positive.

Direction of a Vector

Step 1 of 6
X Y O 1 square = 1 unit θ R(3, 3) M(3, 0) OM = 3 RM = 3 θ = 45°
X Y O θ A(x, y) y M x
X Y O θ P(x₁, y₁) Q(x₂, y₂) y₂ – y₁ x₂ – x₁ R M N

Derivation Steps

In the figure, what is the angle made by the line segment OR with the positive direction of the X-axis? Let this angle be θ.

tanθ = RMOM = 33 = 1 = tan45°

∴  θ = 45°

Here, OR makes an angle of 45° with the positive X-axis. This angle 45° is called the direction of vector OR.

Similarly, let the angle made by OA with the positive X-axis be θ.

tanθ = AMOM = yx

∴  θ = tan–1( yx )

Thus, the direction of OA is θ = tan–1( yx ) degrees.

Likewise, let PQ make an angle θ with the positive X-axis, where P(x1, y1) and Q(x2, y2).

tanθ = QRPR = y2 – y1x2 – x1

∴  θ = tan–1( y2 – y1x2 – x1 )

Thus, the direction of PQ (θ) = tan–1( y2 – y1x2 – x1 ).

The angle made by any vector with the positive direction of the X-axis is called the direction of that vector.

θ = tan–1( Y-componentX-component )

Unit Vector

Step 1 of 5

Derivation Steps

A vector whose magnitude is 1 unit is called a unit vector.

For a = (1, 0), the magnitude is |a| = √1² + 0² = √1 + 0 = √1 = 1

Similarly, for b = (0, 1), the magnitude is |b| = √0² + 1² = √0 + 1 = √1 = 1 unit.

Thus, both a and b have magnitude 1, so they are unit vectors.

If any vector a is divided by its magnitude |a|, the resulting vector is the unit vector in the direction of a. It is denoted by a.

i.e., a = a|a|

Unit Vector Parallel to X-axis and Y-axis

Step 1 of 7
1 square = 1 unit X Y O i (1, 0) j (0, 1)

Derivation Steps

The unit vector parallel to the X-axis is denoted by i = (1, 0), and the unit vector parallel to the Y-axis is denoted by j = (0, 1).

If a = (x, y), then it can be written as

a = xi + yj

a = xi + yj

= x(1, 0) + y(0, 1)

= (x, 0) + (0, y)

= (x + 0, 0 + y)

= (x, y)

Hence a = xi + yj = (x, y).

Example 1 of 2

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