A vector that shows direction and has a magnitude of 1 is called a unit vector. Use these steps to determine the unit vector of a given vector: Determine the vector first. Let v be the vector that needs to have its unit vector found. Assume that v=(v1, v2, v3 ) in three-dimensional space. The next step is to calculate the magnitude (or length) of the vector v. The magnitude of the vector v is given by the formula

∥v∥= √v1² + v2²+ v3²

​ This formula can be adjusted for two-dimensional vectors by excluding the v3 component. Once the magnitude is calculated, the next step is to divide each component of the v by its magnitude. The unit vector u in the direction of v is found by dividing each component of v by its magnitude using the formula

u = (v1/∥v∥ , v2/∥v∥, v3/∥v∥)

Definition of Unit vector

A unit vector is a vector that has a magnitude (or length) of 1. This means that it represents direction only, without any information about quantity or magnitude. In other words, a unit vector is a “pure” direction vector, stripped of any information about how “far” in that direction we might want to go. This makes it a very useful tool in physics and engineering, where we often want to separate the concepts of direction and magnitude.

Importance of unit vectors

• Directional Representation: Unit vectors can be used to represent the direction of a physical quantity. For instance, in physics, we often deal with quantities like force, velocity, and acceleration, which have both magnitude and direction. The direction of these quantities can be conveniently represented using unit vectors.
• Simplification of Calculations: Unit vectors simplify many vector calculations. For example, when you need to find the component of a force vector along a certain direction, you can simply compute the dot product of the force vector and a unit vector in the desired direction.
• Separation of Magnitude and Direction: Unit vectors allow us to separate the magnitude and direction of a vector. This is particularly useful when we want to analyze the directional properties of a vector independently of its magnitude.
• Standardization: Unit vectors provide a standard way of describing directions in space. In Cartesian coordinates, the standard unit vectors i, j, and k represent the x, y, and z directions, respectively.
• Vector Decomposition: Any vector can be expressed as the sum of scaled unit vectors. This is known as the component form of a vector and is a fundamental concept in vector algebra.

Representation of a Unit Vector

Unit vectors are usually represented by a lowercase letter with a hat (or caret) symbol, such as â / Â. This notation distinguishes unit vectors from regular vectors. The hat symbol is a universal notation in mathematics and physics to denote unit vectors. It helps to visually differentiate between vectors of arbitrary magnitude and those of unit magnitude.

Formula for a Unit Vector

The formula for a unit vector in the direction of a given vector is:

â = a / ∣a|​

Here, a is the given vector, |a| is the magnitude of the vector a, and â is the unit vector in the direction of a. This formula essentially scales the original vector down to a magnitude of 1, while preserving its direction.

Calculating Unit Vector

Calculating a unit vector involves two steps: finding the magnitude of the original vector and then dividing the original vector by its magnitude

1. Identify the magnitude of the vector:

The magnitude of a vector is calculated using the Pythagorean theorem. For a 2-D vector a = (x,y) ,

the magnitude is calculated a s |a|=√(x² + y²) .

For a 3-D vector a = (x,y,z) , the magnitude is |a| = √(x² + y² + z²) .

2. Divide the vector by its magnitude:

Once you have calculated the magnitude divide each component of the vector by its magnitude to get the unit vector. Ex : Let a=(x,y,z) then unit vector â = (x / |a| , y/ |a| , z / |a| ) . This operation puts the original vector down to a magnitude of 1 while keeping its direction intact

Sample Problems – How to calculate the Unit Vector

Problem 1: Given a =2 î + 2ĵ + k̂ . Find â

Solution:

Modulus of the vector, |a| = √(x² + y² + z²)

= √(2² + 2² + 1²)= √9

= 3

Unit vector,  â = a / |a|

=  (2 î + 2 ĵ + k̂ ) / 3

= [Tex]{{\frac{2\hat{i}}{3} + \frac{2\hat{j}}{3} + \frac{1\hat{k}}{3}}}[/Tex]

Problem 2: Is a vector given by [Tex]{{\vec{a} = 1\hat{i} + 1\hat{j} + 1\hat{k}}} [/Tex] also a unit vector?

Solution:

Modulus of the vector, [Tex]|a| = \sqrt{{x}^2+{y}^2+{z}^2}[/Tex]

= [Tex]{\sqrt{1^2+1^2+1^2}} [/Tex] = √3

The magnitude of this vector is not 1. Hence, it is not a unit vector.

Problem 3: If [Tex]{{\vec{a}={\frac{1}{4}\hat{i} + \frac{1}{4}\hat{j} + z\hat{k}}}}[/Tex] is a unit vector then find the value of z.

Solution:

The magnitude of a unit vector is 1, which means:

[Tex]|a| = \sqrt{{x}^2+{y}^2+{z}^2} = 1[/Tex]

which means, 

[Tex]{\sqrt{{(1/2)}^2+{(1/2)}^2+{z}^2} = 1}[/Tex]

Squaring both sides,

[Tex]{{{(1/2)}^2+{(1/2)}^2+{z}^2} = 1}[/Tex]

[Tex]{z^2=\frac{1}{2}}[/Tex]

[Tex]{z=\pm{\frac{1}{\sqrt{2}}}}[/Tex]

Question 4: Find unit vector along [Tex]{{\vec{a} = 4\hat{i}}}[/Tex].

Solution:

Modulus of the vector, [Tex]|a| = \sqrt{{x}^2+{y}^2+{z}^2}[/Tex]

= [Tex]{\sqrt{4^2+0^2+0^2}}={4}[/Tex]

Unit vector, [Tex]{\hat{a} = \frac{\vec{a}} {|a|}}[/Tex]

= [Tex]{\frac{{4\hat{i}}}{4}}[/Tex]

= [Tex]{{\hat{i}}}[/Tex]

Question 5: If unit vector along [Tex]{\vec{A}}[/Tex] of magnitude 2√2 is [Tex]{\frac{1\hat{i}}{\sqrt{2}} + \frac{1\hat{j}}{\sqrt{2}}}[/Tex] . Find [Tex]{\vec{A}}[/Tex].

Solution:

Unit vector, [Tex]{\hat{A} = \frac{\vec{A}} {|A|}}[/Tex]

Which means [Tex]{\vec{A} = {\hat{a}}\times{|a|}}[/Tex]

Thus, [Tex]{\vec{A} = {({\frac{1\hat{i}}{\sqrt{2}} + \frac{1\hat{j}}{\sqrt{2}}}})}\times{2\sqrt{2}}[/Tex]

[Tex]{\vec{A} = 2\hat{i}+2\hat{j}}[/Tex]

Question 6: A force of 5 N is acting at an angle of 60 degrees from the positive x-axis in the xy-plane. Calculate the unit vector in the direction of this force.

Solution :

A Force of 5 N at an angle of 60 degrees from x-the axis can be represented as F = (Fx, Fy), where Fx and Fy are components of the force along the x and y axes

First, we would calculate the component of the force using the given magnitude and direction :

Fx = F*cos(60) = 5 * (cos(60)) = 5* (1/2) = 5/2

Fy = F*(sin(60)) = 5* (sin(60)) = 5 *(√3 / 2 ) = 5√3 / 2

So F = (5√3 / 2 N, 5/2N )

Next, we have to calculate the magnitude of the force

|F| = √Fx² + Fy² = (5√3 / 2) ² + (5/2)² = 25 * 3 /4 + 25/4 = √ 100 / 4 =10/2 = 5

So the magnitude of F is 5 N as expected.

Finally the unit vector in the direction of F is = F/|F| = (Fx/|F| , Fy/|F|) = (5/2 * 5 , 5√3 / 2 * 5) =

=(1/2,√3 / 2 )

= (0.5, 0.866)

So the unit vector in the director of F is = (0.5 , 0.866 ) or (0.5 î + 0.866 ĵ)