The dot product is one means of multiplying two or more vectors. The result of the period product of vectors is a scalar quantity. Thus, the dot product is likewise known together a scalar product. Algebraically, the is the amount of the commodities of the matching entries of 2 sequences that numbers. Geometrically, the is the product the the Euclidean magnitude of 2 vectors and the cosine of the angle in between them. The period product the vectors finds various applications in geometry, mechanics, engineering, and also astronomy. Allow us comment on the period product in detail in the upcoming sections.

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1. | What is dot Product of 2 Vectors? |

2. | Dot Product Formula |

3. | Geometrical meaning of period Product |

4. | Matrix Representationof period Product |

5. | Properties ofDot Product |

6. | Dot Product that Unit Vectors |

7. | Application of period Product |

8. | FAQs on dot Product |

## What is dot Product of 2 Vectors?

Dot Product that vectors is same to the product the the magnitudes that the two vectors, and also the cosine of the angle between the two vectors. The resultant of the period product of two vectors lie in the same plane of the 2 vectors. The period product might be a positive real number or a an unfavorable real number.

### Dot Product Definition

In vector algebra, if 2 vectors are given as:a= <\(a_1\),\(a_2\),\(a_3\),\(a_4\),….,\(a_n\)> and b = <\(b_1\),\(b_2\),\(b_3\),\(b_4\),….,\(b_n\)>then their dot product is offered by:a.b = \(a_1 b_1\)+\(a_2 b_2\)+\(a_3 b_3\)+……….\(a_n b_n\)

**\(\overrightarrow a. \overrightarrow b=\sum_i=1^n a_i b_i\)**

## Dot Product Formula because that Vectors

Let a and b be 2 non-zero vectors, and θ be the consisted of angle of the vectors. Climate the scalar product or period product is denoted by a.b, i beg your pardon is defined as:

**\(\overrightarrow a. \overrightarrow b\) = \(|\overrightarrow a||\overrightarrow b|\)cos θ**.

Here, \(|\overrightarrow a|\) is the size of \(\overrightarrow a\),

\(|\overrightarrow b|\) is the size of \(\overrightarrow b\), and θ is the angle in between them.

Note: θ is not identified if one of two people \(\overrightarrow a\) = 0 or \(\overrightarrow b\) = 0.

## Geometrical an interpretation of dot Product

The dot product of two vectors is built by taking the component of one vector in the direction the the other and multiplying it with the magnitude of the other vector. To understand the vector dot product, we an initial need come know just how to uncover the size of two vectors, and also the angle between two vectors to find the forecast of one vector over an additional vector.

### Magnitude the A Vector

A vector to represent a direction and also a magnitude. The magnitude of a vector is the square root of the amount of the squares the the separation, personal, instance constituents of the vector. The size of a vector is a confident quantity. For a vector \(\overrightarrow a = a_1x + a_2y + a_3z\), the size is |a| and is provided by the formula, \(|\overrightarrow a| = \sqrta_1^2 + a_2^2 +a_3^2\)

### Projection the a Vector

The dot product is helpful for detect the component of one vector in the direction that the other. The vector projection of one vector over an additional vector is the length of the zero of the offered vector over an additional vector. It is obtained by multiply the magnitude of the given vectors through the cosecant of the angle in between the two vectors. The resultant of a vector projection formula is a scalar value.

Let OA = \(\overrightarrow a\), OB = \(\overrightarrow b\), it is in the two vectors and θ it is in the angle in between \(\overrightarrow a\) and also \(\overrightarrow b\). Attract AL perpendicular come OB.

From the appropriate triangle OAL , cos θ = OL/OA

OL = OA cos θ = \(|\overrightarrow a|\) cos θ

OL is the vector projection of a on b.

\(\overrightarrow a. \overrightarrow b\) = \(|\overrightarrow a||\overrightarrow b|\)cos θ = \(|\overrightarrow b|\) OL

= \(|\overrightarrow b|\) (projection that \(\overrightarrow a\) ~ above \(\overrightarrow b\))

Thus, estimate of \(\overrightarrow a\) ~ above \(\overrightarrow b = \dfrac\overrightarrow a. \overrightarrow b\)

Similarly, the vector estimate of \(\overrightarrow b\) top top \(\overrightarrow a = \dfrac\overrightarrow a. \overrightarrow b\)

### Angle between Two Vectors making use of Dot Product

The angle between two vectors is calculated together the cosine of the angle between the two vectors. The cosine of the angle between two vectors is same to the sum of the product of the individual constituents the the 2 vectors, split by the product the the size of the two vectors. The formula for the angle in between the two vectors is together follows.

\(cos\theta = \dfrac\overrightarrow a.\overrightarrow b.\)

\(cos\theta = \dfraca_1.b_1 + a_2.b_2 +a_3.b_3\sqrta_1^2 + a_2^2 +a_3^3.\sqrtb_1^2 + b_2^2 + b_3^2\)

### Working preeminence to uncover The dot Product of 2 Vectors

If the two vectors space expressed in terms of unit vectors, i, j, k, along the x, y, z axes, climate the scalar product is acquired as follows:

If \(\overrightarrow a = a_1\hat i + a_2 \hat j + a_3 \hat k\) and \(\overrightarrow b = b_1 \hat i + b_2 \hat j + b_3\hat k\), then

\(\overrightarrow a. \overrightarrow b\) = \((a_1 \hat ns + a_2 \hat j + a_3 \hat k)(b_1 \hat ns + b_2 \hat j + b_3 \hat k)\)

= \((a_1b_1) (\hat i. \hat i) + (a_1b_2) (\hat i.\hat j)+ (a_1b_3) (\hat i. \hat k) + \\(a_2b_1) (\hat j. \hat i) + (a_2b_2)(\hat j. \hat j) + (a_2b_3 (\hat j. \hat k) + \\(a_3b_1)(\hat k. \hat i) + (a_3b_2)(\hat k. \hat j) + (a_3b_3)(\hat k. \hat k)\)

\(\overrightarrow a. \overrightarrow b\) = \(a_1b_1\) + \(a_2b_2\)+ \(a_3b_3\)

It is easy to compute the dot product the vectors if the vectors are stood for as heat or pillar matrices. The transpose procession of the an initial vector is derived as a heat matrix. Procession multiplication is done. The heat matrix and column matrix room multiplied to gain the amount of the product the the corresponding materials of the 2 vectors.

The adhering to are the properties of the period product the vectors.

Commutative propertyDistributive propertyNatural propertyGeneral propertiesVector identities### Commutative property of period Product:

With the usual definition, \(\overrightarrow a\). \(\overrightarrow b\) = \(\overrightarrow b\) . \(\overrightarrow a\) , we have \(|\overrightarrow a||\overrightarrow b|\)cos θ = \(|\overrightarrow b||\overrightarrow a|\)cos θ

### Distributivity of period Product

Let a, b, and c be any kind of three vectors, climate the scalar product is distributive over addition and subtraction. This property deserve to be prolonged to any variety of vectors.

\(\overrightarrow a. (\overrightarrow b+\overrightarrow c) = \overrightarrow a. \overrightarrow b + \overrightarrow a. \overrightarrow c\)\((\overrightarrow a+\overrightarrow b). \overrightarrow c = \overrightarrow a. \overrightarrow c+ \overrightarrow b. \overrightarrow c\)\(\overrightarrow a. (\overrightarrow b - \overrightarrow c) = \overrightarrow a. \overrightarrow b - \overrightarrow a. \overrightarrow c\)\((\overrightarrow a -\overrightarrow b). \overrightarrow c = \overrightarrow a. \overrightarrow c - \overrightarrow b. \overrightarrow c\)### Nature of dot Product

We understand that 0 ≤ θ ≤ π.If θ = 0 then**a**.

**b**= abdominal

**a**.

**b**= -ab

**a**.

**b**= 0

### Other properties of dot Product

Let a and b be any kind of two vectors, and also λ be any type of scalar. Then (λ\(\overrightarrow a) . \overrightarrow b\) = λ (\(\overrightarrow a) \overrightarrow b\)For any two scalars λ and μ, λ\(\overrightarrow a\) . μ \(\overrightarrow b\) = (λμ\(\overrightarrow a)). \overrightarrow b\) = \(\overrightarrow a\). (λμ \(\overrightarrow b\))The length of a vector is the square source of the period product that the vector by itself. \(\overrightarrow a\) = \(\sqrt\overrightarrow a . \overrightarrow a\)\(\overrightarrow a. \overrightarrow a\) = \(|\overrightarrow a\)|2 = (\(\overrightarrow a\))2 = \(\overrightarrow a\)2For any type of two vectors a and also b, \(|\overrightarrow a + \overrightarrow b|\) ≤ |\((\overrightarrow a\)| + |(\(\overrightarrow b|\)### Vector Identities

(\(\overrightarrow a + \overrightarrow b\)) 2 = \(|\overrightarrow a\)|2 + \(|\overrightarrow b|\)2 + 2 \((\overrightarrow a.\overrightarrow b)\)(\(\overrightarrow a - \overrightarrow b\)) 2 = \(|\overrightarrow a\)|2 + \(|\overrightarrow b|\)2 - 2 \((\overrightarrow a.\overrightarrow b)\)\((\overrightarrow a + \overrightarrow b). (\overrightarrow a - \overrightarrow b) = |\overrightarrow a\)|2 - \(|\overrightarrow b|\)2 ≤ \(|\overrightarrow a\)| + \(|\overrightarrow b|\)The dot product the the unit vector is studied by acquisition the unit vectors \(\hat i\) along the x-axis, \(\hat j\) along the y-axis, and \(\hat k\) along the z-axis respectively. The period product the unit vectors \(\hat i\), \(\hat j\), \(\hat k\) follows similar rules as the period product of vectors. The angle in between the same vectors is same to 0º, and hence their period product is equal to 1. And also the angle in between two perpendicular vectors is 90º, and their dot product is same to 0.

\(\hat i.\hat i\) = \(\hat j.\hat j\) = \(\hat k.\hat k\)= 1

\(\hat i.\hat j\) = \(\hat j.\hat k\) = \(\hat k.\hat i\)= 0

The applications of the scalar product is the calculation of work. The product that the pressure applied and also the displacement is dubbed the work. If force is exerted at an edge θ to the displacement, the work done is provided as the dot product the force and also displacement together W = f d cos θ. The period product is additionally used to check if two vectors are orthogonal or not. \(\overrightarrow a. \overrightarrow b\) = \(|\overrightarrow a||\overrightarrow b|\) cos** **90º ⇒ \(\overrightarrow a. \overrightarrow b\) = 0

**Important note on period Product**

☛** additionally Check:**

**Example 2. Uncover the angle between the two vectors 2i + 3i + k, and 5i -2j + 3k.**

**Solution:**

The two offered vectors are:

\(\overrightarrow a\) = 2i + 3i + k, and also \(\overrightarrow b\) = 5i -2j + 3k

|a| = \(\sqrt2^2 + 3^2 + 1^2\) = \(\sqrt4 + 9 + 1\) = \(\sqrt14\)

|b| = \(\sqrt5^2 + (-2)^2 + 3^2\) = \(\sqrt25 + 4 + 9\) = \(\sqrt38\)

Using the dot product we have actually \(\overrightarrow a.\overrightarrow b\) = 2.(5) + 3.(-2) + 1.(3) = 10 - 6 + 3 = 7

Cosθ = \( \dfraca.b\)

= \(\dfrac7\sqrt14.\sqrt38\)

= \(\dfrac72.\sqrt7 \times 19\)

= \(\dfrac72 \sqrt 133\)

θ = Cos-1\(\dfrac72 \sqrt 133\)

θ = Cos-1 0.304 = 72.3°

**Answer:** thus the angle in between the vectors is 72.3°

**Example 3. Check whether vectors a = 3i + 2j -k and b = i - 2j - k space orthogonal. **

**Solution:**

To examine if 2 vectors room perpendicular, we compute the dot product and also find if the an outcome is 0.

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Given: a = 3i + 2j -k and b = ns - 2j - k.

The dot product is computed as \(\overrightarrow a.\overrightarrow b = a_1b_1 + a_2b_2+ a_3b_3\)