Components and Projection of a Vector: Formula, Derivation, Examples

Components and projection of a vector: The components of a vector helps to split a vector into different parts. This helps perform various algebraic operations with the vectors. Components of a vector represent part of the vector with reference to each of the axes of the coordinate system. The projection gives the projection of one vector over another. This is a scalar quantity. Vector projection has many applications in Physics and Engineering for representing a force with respect to another vector.

In this article, let us learn about the vector projection, formula, derivation, and the components of a vector with some solved examples.

What are the Components of a Vector?

Vectors are generally represented in a \(2−\)dimensional coordinate plane, with an \(x−\)axis and \(y−\)axis, or \(3−\)dimensional space, containing the \(x−\)axis, \(y−\)axis, and \(z−\)axis, respectively. The components of a vector help split a given vector. Any vector can be split with reference to the coordinate axes, and we can compute the components of a vector. The individual components of a vector are then combined to obtain the complete vector representation.

In the \(2−\)dimensional coordinate system, the direction of the vector is the angle made by the vector with the positive \(x−\)axis. Let \(Z\) be the vector, and \(\theta \) is the angle made by the vector with the positive direction of \(x−\)axis. So, we have the components of this vector along the \(x−\)axis  and \(y−\)axis as \({Z_x},\) and \({Z_y}\) respectively.

So,
\({Z_x} = {\rm{Z}}\cos \theta \)

\({Z_y} = Z\sin \theta \)

\( \Rightarrow |Z| = \sqrt {Z_x^2 + Z_y^2} \)

Further, the vector is also represented in \(3−\)dimensional space as \(\vec A = a\hat i + b\hat j + c\hat k.\) Here \(\hat i,\hat j,\) and \({\hat k}\) are the unit vectors along \(x−\)axis, \(y−\)axis, and \(z−\)axis, respectively.

The unit vectors help identify the components of the vectors with reference to the coordinate axes.

How to Find Components of a Vector?

The component form of a vector is \(\vec A = a\hat i + b\hat j + c\hat k.\) The values \(a,b,c\) are called scalar components of the vector \(A,\) and \(a\hat i,b\hat j,c\hat k\) are called vector components of the vector \(A.\) Sometimes \(a,b,c\) are called the rectangular components.

Algebraic Operations Using Components of a Vector

If \(\vec a\) and \(\vec b\) are two vectors in the component form \({a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) and \({b_1}\hat i + {b_2}\hat j + {b_3}\hat k,\) respectively, then:

(i) The sum or resultant vectors of \(\vec a\) and \(\vec b\) is given by
\(\vec a + \vec b = \left( {{a_1} + {b_1}} \right)\hat i + \left( {{a_2} + {b_2}} \right)\hat j + \left( {{a_3} + {b_3}} \right)\hat k\)

(ii) The difference between the vectors \(\vec a\) and \(\vec b\)  is given by
\(\vec a – \vec b = \left( {{a_1} – {b_1}} \right)\hat i + \left( {{a_2} – {b_2}} \right)\hat j + \left( {{a_3} – {b_3}} \right)\hat k\)

(iii) Vectors \(\vec a\) and \(\vec b\) are parallel, if and only if \(\vec b = \lambda \vec a = \left( {\lambda {a_1}} \right)\hat i + \left( {\lambda {a_2}} \right)\hat j + \left( {\lambda {a_3}} \right)\hat k\)

What is a Projection Vector?

The projection of a vector is the length of the shadow of the given vector on another vector. It is the product of the magnitude of the given vector and the cosine of the angle between the two vectors. 

Projection Vector Formula

The projection of vector \(\vec a\) on vector \(\vec b\) is equal to the dot product of vector \(\vec a\) and vector \(\vec b,\) divided by the magnitude of the vector \(\vec b.\)

The projection of a vector \(\vec a\) on vector \(\vec b\) is given by

\(\frac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow b } \right|}}\)

Derivation

The following derivation helps us understand the projection vector of one vector over another vector.

Let \(OA = \vec a,OB = \vec b\) be the two vectors and \(\theta \) be the angle between two vectors \(\vec a\) and \(\vec b.\) Draw \(AL\) perpendicular to \(OB.\)

So, from the triangle \(OAL,\) we have 

\(\cos \theta = \frac{{OL}}{{OA}}\)

\( \Rightarrow OL = OA\cos \theta \)

\( \Rightarrow OL = |\vec a|\cos \theta \)

 \(OL\) is the projection of vector \(\vec a\) on vector \(\vec b.\) Then, 

\(\vec a \cdot \vec b = |\vec a||\vec b|\cos \theta \)

\( \Rightarrow \vec a \cdot \vec b = |\vec b|(|\vec a|\cos \theta )\)

\( \Rightarrow \vec a \cdot \vec b = |\vec b|OL\)

\( \Rightarrow OL = \frac{{\vec a \cdot \vec b}}{{|\vec b|}}\)

Hence, the projection of a vector \(\vec a\) on vector \(\vec b\) is  \(\frac{{\vec a \cdot \vec b}}{{|\vec b|}}\)

Similarly, the projection of vector \(\vec b\) on vector \(\vec a\) is \(\frac{{\vec a \cdot \vec b}}{{|\vec a|}}\)

Applications of Projection Vectors

(i) Dot Product of Two Vectors:

The dot product of two vectors is a scalar product, if two vectors are expressed in terms of unit vectors \(\hat i,\hat j,\) and \({\hat k},\) along the \(x−\)axis, \(y−\)axis, and \(z−\)axis.

Let \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) and \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) be the two non-zero vectors, then 

\(\vec a \cdot \vec b = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right) \cdot \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right)\)

\( \Rightarrow \vec a \cdot \vec b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\)

(ii) Finding the Angle Between Two Vectors:

Let \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) and \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) be the two non-zero vectors, then the angle between two vectors is given by

\(\cos \theta = \frac{{\vec a \cdot \vec b}}{{|\vec a||\vec b|}}\)

\(\therefore \cos \theta = \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{\sqrt {a_1^2 + a_2^2 + a_3^2} \cdot \sqrt {b_1^2 + b_2^2 + b_3^2} }}\)

(iii) Finding Components of \(\vec b\) Along and Perpendicular to \(\vec a\):

Let \(\vec a\) and \(\vec b\) be two vectors represented by \(\overrightarrow {OA} \) and \(\overrightarrow {OB} \) respectively, let \(\theta \) be the angle between \(\vec a\) and \(\vec b\). Then \(\vec b = \overrightarrow {OM} + \overrightarrow {MB} \)  [using triangle law of addition]

Also, \(\overrightarrow {OM} = (OM)\hat a\)

 \( = (OB\cos \theta )\hat a\)

 \( = (|\vec b|\cos \theta )\hat a\)

 \( = \left( {|\vec b|\frac{{(\vec a \cdot \vec b)}}{{|\vec a||\vec b|}}} \right)\hat a\)

 \( = \left( {\frac{{\vec a \cdot \vec b}}{{|\vec a|}}} \right)\hat a\)

 \( = \frac{{\vec a \cdot \vec b}}{{|\vec a||\vec a|}}\vec a\)

 \( = \left( {\frac{{\vec a \cdot \vec b}}{{|\vec a{|^2}}}} \right)\vec a\)

Similarly, \(\overrightarrow {MB} = \vec b – \overrightarrow {OM} = \vec b – \left( {\frac{{\vec a \cdot \vec b}}{{|\vec a{|^2}}}} \right)\vec a\)

Thus, the components of \(\vec b\) along and perpendicular to \(\vec a\) are \(\left( {\frac{{\vec a \cdot \vec b}}{{|\vec a{|^2}}}} \right)\vec a\) and \(\vec b – \left( {\frac{{\vec a \cdot \vec b}}{{|\vec a{|^2}}}} \right)\vec a,\) respectively.

Solved Examples

Q.1. Find the projection of a vector \(4\hat i + 2\hat j + \hat k\) on a vector \(5\hat i – 3\hat j + 3\hat k.\)
Ans: Let \(\vec a = 4\hat i + 2\hat j + \hat k\) and \(\vec b = 5\hat i – 3\hat j + 3\hat k\)
As we know, the projection of a vector \({\vec a}\) on vector \({\vec b}\) is \(\frac{{\vec a \cdot \vec b}}{{|\vec b|}}\)
Therefore, the projection of a vector \(4\hat i + 2\hat j + \hat k\) on a vector
\(5\hat i – 3\hat j + 3\hat k = \frac{{(4\hat i + 2\hat j + \hat k) \cdot (5\hat i – 3\hat j + 3\hat k)}}{{|5\hat i – 3\hat j + 3\hat k|}}\)
\( = \frac{{4 \cdot (5) + 2( – 3) + 1(3)}}{{\sqrt {{5^2} + {{( – 3)}^2} + {{(3)}^2}} }}\)
\( = \frac{{20 – 6 + 3}}{{\sqrt {25 + 9 + 9} }}\)
\( = \frac{{17}}{{\sqrt {43} }}\)

Q.2. Find the projection of a vector \(5\hat i + 4\hat j + \hat k\) in the direction of the vector \(3\hat i + 5\hat j – 2\hat k.\)
Ans: Let \(\vec a = 5\hat i + 4\hat j + \hat k\) and \(\vec b = 3\hat i + 5\hat j – 2\hat k\)
As we know, the projection of a vector \({\vec a}\) on vector \({\vec b}\) is \(\frac{{\vec a \cdot \vec b}}{{|\vec b|}}\)
Therefore, the projection of a vector \(5\hat i + 4\hat j + \hat k\) on a vector
\(3\hat i + 5\hat j – 2\hat k = \frac{{(5\hat i + 4\hat j + \hat k) \cdot (3\hat i + 5\hat j – 2\hat k)}}{{|3\hat i + 5\hat j – 2\hat k|}}\)
\( = \frac{{5 \cdot (3) + 4(5) + 1( – 2)}}{{\sqrt {{3^2} + {{(5)}^2} + {{( – 2)}^2}} }}\)
\( = \frac{{15 + 20 – 2}}{{\sqrt {9 + 25 + 4} }}\)
\( = \frac{{33}}{{\sqrt {38} }}\)

Q.3. Find the component of \(\vec a = 2\hat i + 3\hat j\) along the direction of \(\hat i + \hat j.\)
Ans: Given that, \(\vec a = 2\hat i + 3\hat j\)
Let \(\vec b = \hat i + \hat j\)
So, the unit vector of \({\vec b}\) is \(\hat b = \frac{{\hat i + \hat j}}{{\left| {\hat j + \hat j} \right|}} = \frac{1}{{\sqrt 2 }}(\hat i + \hat j)\)
The component of a vector \(\vec a\) along the direction of the vector \(\vec b = a\cos \theta \)
\( = (\vec a \cdot \hat b)\hat b\)
\( = \left[ {(2\hat i + 3\hat j) \cdot \frac{1}{{\sqrt 2 }}(\hat i + \hat j)} \right]\frac{1}{{\sqrt 2 }}(\hat i + \hat j)\)
\( = \left[ {\frac{1}{{\sqrt 2 }}(2 + 3)} \right]\frac{1}{{\sqrt 2 }}(\hat i + \hat j)\)
\( = \frac{5}{2}(\hat i + \hat j)\)

Q.4. Find the component of a vector \(2\hat i + 3\hat j + 2\hat k\) perpendicular \(\hat i + \hat j + \hat k\)
Ans: Let \(\vec a = 2\hat i + 3\hat j + 2\hat k\) and \(\vec b = \hat i + \hat j + \hat k\)
As we know, the component of vector \({\overrightarrow a }\) perpendicular to vector \({\overrightarrow b }\) is given by \(\vec a – \frac{{\vec a \cdot \vec b}}{{|\vec b{|^2}}} \times \vec b\)
\( = 2\hat i + 3\hat j + 2\hat k – \frac{{(2\hat i + 3\hat j + 2\hat k)(\hat i + \hat j + \hat k)}}{{(\sqrt {1 + 1 + 1} }} \times (\hat i + \hat j + \hat k)\)
\( = 2\hat i + 3\hat j + 2\hat k – \frac{1}{{{{(\sqrt 3 )}^2}}} \times (\hat i + \hat j + \hat k)\)
\( = \frac{5}{3}(\hat i – 2\hat j + \hat k)\)
Hence, the component of a vector \({\vec a}\) is perpendicular to vector \({\vec b}\) is \(\frac{5}{3}(\hat i – 2\hat j + \hat k).\)

Q.5. If \(\vec a = 2\hat i – 3\hat j + 6\hat k\) and \(\vec b = – 2\hat i – 2\hat j + \hat k,\) then find \(\frac{{{\rm{Projection}}\,{\rm{of}}\,{\rm{vector}}\,\vec a\,{\rm{on}}\,{\rm{vector}}\,\vec b}}{{{\rm{Projection}}\,{\rm{of}}\,{\rm{vector}}\,\vec b\,{\rm{on}}\,{\rm{vector}}\,\vec a}}.\)
Ans: Given that, we have: \(\vec a = 2\hat i – 3\hat j + 6\hat k\) and \(\vec b = – 2\hat i – 2\hat j + \hat k\)
As we know, the projection of a vector \(\vec a\) on vector \(\vec b\) is \(\frac{{\vec a \cdot \vec b}}{{|\vec b|}}\)
Therefore, the projection of a vector \(2\hat i – 3\hat j + 6\hat k\) on a vector
\( – 2\hat i – 2\hat j + \hat k = \frac{{(2\hat i – 3\hat j + 6\hat k) \cdot ( – 2\hat i – 2\hat j + \hat k)}}{{| – 2\hat i – 2\hat j + \hat k|}}\)
\( = \frac{{2 \cdot ( – 2) – 3( – 2) + 6(1)}}{{\sqrt {{{( – 2)}^2} + {{( – 2)}^2} + {{(1)}^2}} }}\)
\( = \frac{{ – 4 + 6 + 6}}{{\sqrt 9 }}\)
\( = \frac{8}{3}\)
Similarly, the projection of a vector \( – 2\hat i – 2\hat j + \hat k\) on a vector
\(2\hat i – 3\hat j + 6\hat k = \frac{{(2\hat i – 3\hat j + 6\hat k) \cdot ( – 2\hat i – 2\hat j + \hat k)}}{{|2\hat i – 3\hat j + 6\hat k|}}\)
\( = \frac{{2( – 2) – 3( – 2) + 6(1)}}{{\sqrt {{{(2)}^2} + {{( – 3)}^2} + {{(6)}^2}} }}\)
\( = \frac{{ – 4 + 6 + 6}}{{\sqrt {4 + 9 + 36} }}\)
\( = \frac{8}{7}\)
Hence,
\(\frac{{{\rm{Projection}}\,{\rm{of}}\,{\rm{vector}}\,\vec a\,{\rm{on}}\,{\rm{vector}}\,\vec b}}{{{\rm{Projection}}\,{\rm{of}}\,{\rm{vector}}\,\vec b\,{\rm{on}}\,{\rm{vector}}\,\vec a}} = \frac{{\frac{8}{3}}}{{\frac{8}{7}}} = \frac{7}{3}\)

Summary

The individual components of a vector are combined to get the entire vector representation. The vector \(\vec A = a\hat i + b\hat j + c\hat k\)  is the component form. The values \(a, b, c\) are called scalar components of a vector \(A,\) and \(a\hat i,b\hat j,c\hat k\) are the vector components of vector \(A.\) The projection of a vector on a vector is the length of the shadow of the given vector on another vector. The projection of a vector \({\vec a}\) on vector \({\vec b}\) is  \(\frac{{\vec a \cdot \vec b}}{{|\vec b|}}\) and the projection of vector \({\vec b}\) on vector \({\vec a}\) is \(\frac{{\vec a \cdot \vec b}}{{|\vec a|}}.\) The applications of projection of a vector are the dot product of vectors, the angle between two vectors, and the components of \({\vec b}\) along and perpendicular to \({\vec a}.\)

Frequently Asked Questions (FAQs)

Q.1. What is the difference between projection and component?
Ans: Although they are basically the same, there is one major difference: ‘reference’. The component is a projection on an axis of the ‘reference’ frame, and projection is a component of a vector along another vector.

Q.2. How do you find the components of a vector along another vector?
Ans: The components of \(\vec b\) along another vector \(\vec a\) is \(\left( {\frac{{\vec a \cdot \vec b}}{{|\vec a{|^2}}}} \right)\vec a\)

Q.3. What is the projection vector formula?
Ans: The projection of a vector \(\vec B\) on vector \(\vec A\) is \(\frac{{\vec B \cdot \vec A}}{{|\vec A|}}.\) This means the projection vector \(\vec B\) on the vector \(\vec A\) is equal to the dot product of two vectors and divided by the magnitude of the vector \(\vec A.\)

Q.4. What is required to find the projection vector?
Ans: We must know the vector for which the projection vector is to be calculated with respect to another vector. To find the projection vector, we must have two vectors followed by the dot product of two vectors and the magnitude of a vector.

Q.5. What is a projection of a vector?
Ans: The projection vector is the vector produced when one vector is resolved into two component vectors, one that is parallel to the second vector and another one perpendicular to the second vector.

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