Unit Vector: Definition, Formula and Examples

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Unit Vectors are those vectors whose magnitude is equal to 1 or one unit, their direction can be anywhere but their magnitude shall remain unity i.e., 1 no matter which direction the unit vector is in. The unit vector in the direction of a given vector r is denoted as r.

Table of Content

Vectors
Magnitude of Vectors
Unit Vectors
Forms of Unit Vector
Things to Remember
Sample Questions

Key Terms: Line segment, Vectors, Magnitude, Co-ordinates, scalar quantities, three-dimensional Cartesian, Unit Vector, Magnitude of vector

Vectors

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Vectors are geometric entities that possess both: magnitude as well as direction as opposed to scalar quantities that only have magnitude. A vector can be pictured as a directed line segment where the length of that line segment is the magnitude of that vector while the arrow in that line segment denotes the direction of the vector, an arrow runs through the whole of this line segment from tail to the head and in whichever direction the arrow is pointing shall be the direction of the vector.

Vectors are written as r or AB (denoting both points of a line segment) and the magnitude of vector r shall be represented as |r| and the magnitude of the vector AB shall be denoted as |AB |.

Interestingly, if a vector is replicated parallel to the pre-existing vector while their length is equal in magnitude and they both have the same direction then they shall be the same vector.

Read More: Co Planar Vectors

Magnitude of Vectors

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The magnitude of the vector is essentially the length of the directed line segment of which the vector is made up. However, a zero vector is such a vector that does not have magnitude i.e., its magnitude is equal to zero and for this vector the direction is undefined. In a Cartesian plane, the magnitude of a vector can be determined as follows:

Let’s assume there is a line segment on the XY plane with two points A and B. so, they make up a vector called AB.

The coordinates of these points are: A (a, b) and B (m, n)

So, the length of this vector AB shall be the length of the line from point A to point B, and using the formulas of distance this length or the magnitude of the vector can be calculated as:

|AB | = \(\sqrt{(m-a)^2 + (n-b)^2}\)

If the vector has one point at the origin and the point has coordinates (a, b) then the formula shall be: |AB | = \(\sqrt{a^2 + b^2}\)

Read More: Types of Vector

Unit Vectors

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Simply put, Unit vectors are vectors with a magnitude of one unit. In other words, the directed line segment that these unit vectors are made of has a length of only one unit. They are denoted by a (the hat or the cap is the symbol of denotation, the alphabet is subject to change).

Read More: Addition of Vectors

Mathematically, vector a is a unit vector-only when |a | = 1 and if it is so, then the vector shall be written as a.

So, for a given vector A its a unit vector (A ) will have the same direction as A but the magnitude will be one regardless of the magnitude of A and the relation between them can be formulated as:

A = (1 / |A |) A

Read More: Vector product of two vectors

Forms of Unit Vector

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The unit vector in three dimensions

In a three-dimensional Cartesian Plane, the following are the unit vectors:

i for the x-axis (indicates the direction of an object along the x-axis.)

j for the y-axis (indicates the direction of an object along the y-axis)

k for z-axis (indicates the direction of an object along the z-axis)

Every vector placed in this three-dimensional Cartesian plane can be expressed as a linear combination of the above-mentioned unit vectors. For example, AB = ai + bj + ck (where i, j, k are the unit vectors of co-ordinate axes)

Read More: Position Vector

Unit normal vector

A normal vector can be defined as a vector that is perpendicular to a surface at a defined point. The unit normal vector is acquired after normalizing the vector and it is also known as “unit normal”. To obtain a unit normal we divide the non-zero normal vector by its vector norm/magnitude.

Read More: Displacement vector

Things to Remember

Any vector can become a unit vector by dividing the vector by its own magnitude.
Unit vector = vector / vector’s magnitude.
A unit vector is also known as a direction vector.
Since the length is never negative, the notation |r| < 0 has no mathematical worth whatsoever.
Any vector can be represented in space using the unit vector.
The dot product of orthogonal unit vectors and cross product of parallel unit vectors is always zero.

Also Read:

Related Articles
Direction cosines	Negative of a vector	resultant vector formula
Triangle law of vector addition	Properties of vector addition	Properties of scalar product of two vectors
Laws of Vector Addition	Vector Formula	Vector Projection Formula
Scalar triple product of vectors	Direction of a vector	Components of vector
Vector Calculus	Vector Space	Multivariable calculus

Sample Questions

Ques: What is a unit normal? (1 mark)

Ans: A unit normal is another name for unit normal vectors which are acquired after normalizing the given vector.

Ques: How do you solve for unit vectors? (1 mark)

Ans: In order to solve for a unit vector in the same direction as that of a given vector, one must divide the vector by its own magnitude.

Ques: What are the applications of unit vectors? (1 mark)

Ans: Unit vectors can be used to represent any vector in the two as well as a three-dimensional plane in the form of its components. They also specify the direction of a vector.

Ques: Can unit vectors be collinear? (1 mark)

Ans: Yes, two or more unit vectors will be called collinear if their cross product is zero.

Ques: Find the unit vector of 2i + j + 4k. (2 mark)

Ans: Let’s assume the given vector to be R.

So, R = 2i + j +4k

|R | = √2² + 1² + 4² = √4 + 1 + 16 = √21

Unit vector of R shall be denoted as R.

R = [ 1/ |R |] R

R = [ 1/ √21] 2i + j + 4k

Hence the unit vector of R is 121 (2i + j +4k).

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Unit Vector: Definition, Formula and Examples

Vectors

Magnitude of Vectors

Unit Vectors

Forms of Unit Vector

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

2.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

3.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

4.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

5.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

6.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

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Unit Vector: Definition, Formula and Examples

Vectors

Magnitude of Vectors

Unit Vectors

Forms of Unit Vector

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

2. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

3. Find: The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

4.Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

5.Find:Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

6.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

2.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

3.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

4.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

5.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

6.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).