## What is Unit Vector?

Vector units have both direction and magnitude. However, sometimes one is interested only in direction and not the magnitude. In such a case, vectors are often considered unit length. These unit vectors are generally used to represent direction, with a scalar coefficient providing the magnitude. A vector decomposition can be expressed as a sum of unit vector and scalar coefficients.

Vector units are often used to represent quantities in Physics such as force, acceleration, quantity, or torque. In this article, we will discuss how to find unit vectors.

### Unit Vector in Physics

The unit vector in physics is a vector of unit magnitude and particular direction. A unit vector determines the only direction. They do not have dimensions and units.

In a rectangular coordinate system, the x-axis, y-axis, and z-axis are represented.

\[\hat{i}\] , \[\hat{j}\] and \[\hat{k}\]

These unit vectors are perpendicular to each other.

丨\[\hat{i}\]丨= 丨\[\hat{j}\]丨=丨\[\hat{k}\]丨= 1

### Unit Vector Definition

Given a vector \[\vec{V}\], one may want to know the vector parallel to \[\vec{V}\] with unit length. In other words, one may be looking to find “ scalar vector” a such that || a \[\vec{V}\] || = 1. As, || a \[\vec{V}\] || is equals to a || \[\vec{V}\] ||, it follows trivially that a = 1 / || \[\vec{V}\] ||

Therefore, a given vector \[\vec{V}\], the unit vector \[\hat{V}\] is defined by

Unit Vector \[\hat{V}\] = \[\vec{V}\] / || \[\vec{V}\] ||

has a unit vector and parallel to \[\vec{V}\]. A unit vector is denoted with a symbol (‘^’) to indicate that it is of unit length.

### Unit Vector Formula

As we know, vectors have both magnitude and direction. They are represented by an arrow \[\vec{a}\]. \[\hat{a}\] represents a unit vector. Any vector can be easily converted into a unit vector by dividing it by vector magnitude. Generally, x, y, and z coordinates are used to write any vectors.

It can be represented in two ways:

\[\vec{a}\] = ( x,y,z) coordinates using the brackets

\[\vec{a}\] = x\[\hat{i}\] + y\[\hat{j}\] + z\[\hat{k}\]

The magnitude of a vector formula given as:

|\[\vec{a}\]| = \[\sqrt{x^{2} + y^{2} + z^{2}}\]

Unit Vector = \[\frac{Vector}{\text{Vector’s Magnitude}}\]

### Example:

### 1. Find Unit Vector \[\hat{A}\] in the Direction of \[\vec{A}\] = ⟨4,3⟩

Using the Pythagorean theorem,

\[\hat{A}\] = \[\sqrt{4^{2} + 3^{2}}\] = \[\sqrt{25}\] = 5

Substituting the values, we get

|\[\vec{A}\]| = 5 → \[\hat{A}\] = \[\vec{A}\]/5 = (\[\frac{4}{5}\], \[\frac{3}{5}\])

### 2. Find Unit Vector \[\hat{A}\] in the Direction of \[\vec{A}\] = ⟨-12,5⟩

Using the Pythagorean theorem,

\[\hat{A}\] = \[\sqrt{12^{2} + 5^{2}}\] = \[\sqrt{144 + 25}\] = \[\sqrt{169}\] = 13

|\[\vec{A}\]| = 13 → \[\hat{A}\] = \[\vec{A}\]/5 = (\[\frac{-12}{13}\], \[\frac{5}{13}\])

### Spherical Coordinate Unit Vector

The unit vectors in spherical coordinate systems are defined as the function of position. It is convenient to express spherical coordinate unit vectors in terms of rectangular coordinate systems which are not themself the function of position.

\[\hat{r}\] = \[\frac{\hat{r}}{r}\] = \[\frac{x\hat{x} + y\hat{y} + z\hat{z}}{r}\] = \[\hat{x}\] sinθ cosØ + \[\hat{y}\]sinθ sinØ - \[\hat{z}\]cosθ

\[\frac{\hat{z \times r}}{Sin \theta}\] = - \[\hat{x}\] sinØ + \[\hat{y}\] sinØ

\[\hat{\theta}\] = \[\hat{\varnothing }\] x \[\hat{r}\] = \[\hat{x}\]cos θ cos Ø + \[\hat{y}\]cos θ sin Ø - \[\hat{z}\] sin θ

### Unit Tangent Vector

Considering a smooth vector-valued function \[\vec{V}\](t), any vector parallel to \[\vec{V}\]’(t₀) is considered as the tangent to the graph of \[\vec{V}\](t) at t = t₀. It is generally used to represent direction \[\vec{V}\]’(t), not magnitude.

Hence, we are considering the unit vector in the direction of \[\vec{V}\]’(t). This gives the unit tangent vector definition as follows:

Let us consider \[\vec{V}\]’(t) as a smooth function on the open interval I. Then, the unit tangent vector \[\hat{t}\](t) in this case is:

\[\hat{t}\](t) = \[\vec{V}\]’(t) / |\[\vec{V}\]’(t)|

### Unit Normal Vector

Let \[\vec{V}\](t) be a vector-valued function where the unit tangent vector, \[\hat{t}\](t) is smooth on an open interval I. The unit normal vector \[\hat{n}\](t) is defined as:

\[\hat{n}\](t) = \[\hat{t}\]’(t)/ | \[\hat{t}\]’(t)|

It is also known as the principal unit vector.

The principal unit normal vector will always look at the way the curve is curving “inside”.

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### Orthogonal Unit Vector

The unit vector along the direction of the orthogonal axis such as the x-axis, y-axis, and z-axis is known as orthogonal unit vectors. Orthogonal unit vectors are represented by \[\hat{i}\], \[\hat{j}\], and \[\hat{k}\]

[Image will be Uploaded Soon]

### How to find Unit Vectors Perpendicular to Two Vectors?

Let us learn to find the unit vector perpendicular to two vectors “ a” and “b”.

The unit vector perpendicular to two vectors a and b are:

± \[\vec{n}\] = ± (\[\vec{a}\] x \[\vec{b}\]) / |\[\vec{a}\] x \[\vec{b}\]|

Vector of the magnitude , perpendicular to both vector a and b are,

± λ \[\vec{n}\] = ± λ(\[\vec{a}\] x \[\vec{b}\]) / |\[\vec{a}\] x \[\vec{b}\]|

### Finding Unit Vector Perpendicular to Two Vectors Example Problem:

Find the unit vector perpendicular to two vectors that are x vector + y vector and x vector - y vector where x vector is equal to i vector + j vector + k vector and y vector is equal to i vector + 2 j vector + 2 k vector.

Solution:

Let x vector + y vector = K vector

Let x vector - y vector = K₁ vector

K vector = ( i + j + k) + ( i + 2j + 2k)

K vector = 2i + 3j + 3k……….(1)

K₁ vector = ( i + j + k) - ( i + 2j + 2k)

K₁ vector = - j - 2k…….(2)

Unit perpendicular vector to both K vector and K₁ vector are:

± (k vector x K₁ vector)/|k vector x K₁ vector|

\[\vec{k}\] x \[\vec{k_{1}}\] = \[\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 2 & 3 & 4\\ 0 & -1 & -2 \end{vmatrix}\]

K x K₁ = i[-6 + 4] - j[-4 - 0] + k[-2 + 0]

K x K₁ = -2i + 4j - 2k

|K x K₁| = \[\sqrt{(-2)^{2}}\] + 4² + (-2)² = \[\sqrt{4 + 16 + 4}\] = \[\sqrt{24}\]

= \[\frac{- 2i + 4j - 2k}{\sqrt{24}}\]

= ± \[\frac{2(- i - 2j + k)}{2\sqrt{6}}\]

= ± \[\frac{- i - 2j + k}{\sqrt{6}}\]

### Facts to Remember

Unit vectors are represented similarly as a normal vector, but with a mark over the letter (example, \[\hat{V}\] being the unit vector of V.

To convert a vector into a unit vector, simply divide it by length \[\vec{V}\] = V/||V||. The result obtained will be in a similar direction as the original vector.

1. What is a Unit Vector in Physics?

A unit vector also known as a directional vector is a vector with a magnitude of 1 unit. It is denoted using a lowercase letter with a cap symbol (‘^’) along with it. Any vector can be easily converted into a unit vector by dividing it by vector magnitude as given below.

Unit Vector = Vector / Vector’s Magnitude

2. Does the Unit Vector Have Any Units?

No, the unit vector has only directions but no unit or dimensions.

3. Can Unit Vectors be Negative?

Two vectors are considered equivalent if they represent the same direction and magnitude. Similar to the scalar which can be either positive or negative, the vectors can also be negative or positive.