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Vector Projection Formula

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Vector Projection is basically the shadow of a vector over another vector. The projection vector is obtained by multiplying the vector with the Cos of the angle between the two vectors. A vector is one which has both magnitude and direction. Two vectors are said to be equal if they have the same magnitude as well as the direction. Vector Projection is essential in solving numerical in physics and mathematics.

In this article, we will learn about what is vector projection, the formula of vector projection, its derivation, and some other related concepts in detail.

What is Vector Projection?

Vector Projection is a method of rotating a vector and placing it on a second vector. Hence, a vector is obtained when a vector is resolved into two components, parallel and perpendicular. The parallel vector is called the Projection Vector. Thus, the Vector Projection is the length of the shadow of a vector over another vector. The vector projection of a vector is obtained by multiplying the vector with the Cos of the angle between the two vectors. Let’s say we have two vectors ‘a’ and ‘b’ and we have to find the projection of the vector a on vector b then we will multiply the vector ‘a’ with cosθ where θ is the angle between vector a and vector b.

Vector Projection Formula

If \vec A        is represented as A and \vec B        is represented as B, the Vector Projection of A on B is given as the product of A with Cosθ where θ is the angle between A and B. The other formula for Vector Projection of A on B is given as the product of A and B divided by the magnitude of B. The Projection Vector obtained so is a scalar multiple of A and has a direction in the direction of B.

Projection-of-Vector-a-on-b

Derivation of Vector Projection Formula

The derivation of Vector Projection is discussed below:

Let us assume, OP = \vec A        and OQ = \vec B        and the angle between OP and OQ is θ. Drawn PN perpendicular to OQ.

In the right triangle OPN, Cos θ = ON/OP

⇒ ON = OP Cos θ

⇒ ON = |\vec A        | Cos θ

ON is the projection vector of \vec A        on \vec B

\vec A.\vec B = |\vec A||\vec B|cos \theta

⇒ \vec A.\vec B = |\vec B(|\vec A||cos \theta)

⇒ \vec A.\vec B = |\vec B|ON

⇒ ON = \frac{\vec A.\vec B}{|\vec B|}

Hence, the ON = |\vec A|.\hat B

Thus the Vector Projection of \vec A        on \vec B        is given as \frac{\vec A.\vec B}{|\vec B|}

the Vector Projection of \vec B        on \vec A        is given as \frac{\vec A.\vec B}{|\vec A|}

Vector Projection Important Terms

To find the vector projection we need to learn to find the angle between two vectors and also to calculate the dot product between two vectors.

Angle Between Two Vectors

The angle between the two vectors is given as the inverse of the cosine of the dot product of two vectors divided by the product of the magnitude of two vectors.

Let’s say we have two vectors \vec A        and \vec B        angle between them is θ

⇒ cos θ = \frac{\vec A.\vec B}{|A|.|B|}

⇒ θ = cos-1\frac{\vec A.\vec B}{|A|.|B|}

Dot Product of Two Vectors

Let’s say we have two vectors \vec A        and \vec B        defined as \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        then dot product between them is given as

\vec A.\vec B = (a_1\hat i + a_2\hat j + a_3\hat k)(b_1\hat i + b_2\hat j + b_3\hat k)

⇒ \vec A.\vec B         = a1b1 + a2b2 +a3b3

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Examples on Vector Projection

Example 1. Find the projection of the vector 4\hat i + 2\hat j + \hat k  and 5\hat i -3\hat j + 3\hat k     .

Solution:

Here, \vec{a}=4\hat i + 2\hat j + \hat k \\\vec{b}=5\hat i -3\hat j + 3\hat k             .

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(4.(5) + 2(-3) + 1.(3))}{|\sqrt{5^2 + (-3)^2 + 3^2}|}\\=\dfrac{17}{\sqrt{43}}

Example 2. Find the projection of the vector 5\hat i + 4\hat j + \hat k              and 3\hat i + 5\hat j - 2\hat k

Solution:

Here, \vec{a}=5\hat i + 4\hat j + \hat k \\\vec{b}=3\hat i + 5\hat j - 2\hat k.

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(5.(3) + 4(5) + 1.(-2))}{|\sqrt{3^2 + 5^2 + (-2)^2}|}\\=\dfrac{33}{\sqrt{38}}

Example 3. Find the projection of the vector 5\hat i - 4\hat j + \hat k           and 3\hat i - 2\hat j + 4\hat k

Solution:

Here, \vec{a}=5\hat i - 4\hat j + \hat k \\\vec{b}=3\hat i - 2\hat j + 4\hat k.

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(5.(3) + ((-4).(-2)) + 1.(4))}{|\sqrt{3^2 + (-2)^2 + (4)^2}|}\\=\dfrac{49}{\sqrt{29}}

Example 4. Find the projection of the vector 2\hat i - 6\hat j + \hat k and 8\hat i - 2\hat j + 4\hat k .

Solution:

Here, \vec{a}=2\hat i - 6\hat j + \hat k \\\vec{b}=8\hat i - 2\hat j + 4\hat k

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(2.(8) + ((-6).(-2)) + 1.(4))}{|\sqrt{8^2 + (-2)^2 + (4)^2}|}\\=\dfrac{32}{\sqrt{84}}

Example 5. Find the projection of the vector 2\hat i - \hat j + 5\hat k and 4\hat i - \hat j + \hat k.

Solution:

Here, \vec{a}=2\hat i - \hat j + 5\hat k \\\vec{b}=4\hat i - \hat j + \hat k.

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(2.(4) + ((-1).(-1)) + 5.(1))}{|\sqrt{4^2 + (-1)^2 + (1)^2}|}\\=\dfrac{14}{\sqrt{18}}

Vector Projection – FAQs

1. Define Projection Vector.

The Projection Vector is the shadow of a vector on another vector.

2. What is the Formula for Projection of Vector?

The Formula for Projection of Vector is given as \frac{\vec A.\vec B}{|\vec B|}

3. How to Find Projection Vector?

The projection vector is found by calculating the dot product of the two vectors divided by the on which the shadow is cast.

4. What are Concepts Required to Calculate Projection Vector?

We need to know the angle between two vectors and dot product of two vectors to calculate vector projection.

5. Where is Projection Vector Used?

Projection Vector is used to solve various physics numerical that require the vector quantity to be split into its components.



Last Updated : 05 Oct, 2023
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