Dot Product Formula: Meaning, Properties & Solved Examples

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Jasmine Grover

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Dot Product can be described as the scalar product of two vectors. A dot product can be defined as a scalar number obtained by performing a specific operation on the vector components. We can apply the dot product only for pairs of vectors that have the same number of dimensions. The dot product is symbolized by a heavy dot. The concept of the dot product is popularly used in Mathematics and Physics.

Key Terms: Dot Product, Scalar Product, Vectors, Cross Product, Vector Product, Commutative Property, Associative Property


Products of Two Vectors

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The multiplication between the two vector quantities are of two types. The first one is the dot product or scalar product and the second one is the vector product or cross product. The result of a dot product is always a scalar, whereas, the result of a cross product is always a vector. These two types of vector multiplication have various applications in geometry, mechanics, and engineering. We will be discussing the dot product.

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Dot Product Formula

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The dot product or the scalar product of two non-zero vectors a and b is defined to be the scalar:

Vector a . Vector b = |a| |b| cosθ = ab cos θ

Where a is the magnitude of vector a, b is the magnitude of vector b, and θ is the angle between the two vectors and its value ranges from 0 to 180 degrees. With the same concept, the dot product of any of the two unit vectors is equal to the cosine of the angle between their directions.

Dot Product Formula

Dot Product Formula

Read More: Scalar Triple Product of Vectors


Important Points of Dot Product

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  • It is called a dot product because of the dot that is used as a symbol between the two vectors. 
  • It is also called a scalar product because of the magnitude of vector a, vector b, and the value of cosine θ, all of which are scalar values. 
  • Sign of the dot product: If vector a and vector b are two non-zero vectors, then their dot product is positive, negative, or zero, as per the angle θ between them is acute, obtuse, or right.
  • Square of a vector: If b is a vector then by the concept of the dot product, vector b . vector b = vector b2. Here, the scalar product of vector b with itself will be 

|b| |b| cos 0 = b.b.1 = b2

  • The dot product of two non-zero vectors is calculated as zero if they are perpendicular to each other

Vector a .Vector b = 0 if a =0 or b = 0 or a is perpendicular to b

Read More: Multiplication of Vectors with Scalar


Angle Between Two Vectors in Terms of Dot Product

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Let Vector a and Vector b are two non-zero vectors. These two vectors can be shifted parallel to themselves so that they can intersect at a point. Then the angle θ between them is called the angle between Vector a and Vector b. The mathematical interpretation is given below:

Cos θ = (Vector a . Vector b) / ab θ = cos-1 (Vector a . Vector b)

Read More: Angle Between Two Vectors


Properties of the Dot Product

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The properties of the dot product are listed as follows:

  • The dot product of two vectors is commutative, i.e., (Vector a . Vector b) = (Vector b . Vector a)
  • For any two vectors a and b, Vector a . (-Vector b) = - (Vector a . Vector b) and (- Vector a) . (- Vector b) = (Vector a . Vector b)
  • The dot product of two vectors distributes the addition of vectors, i.e., Vector a . (Vector b + Vector c) = Vector a . Vector b + Vector a . Vector c
  • The dot product is associative with respect to a scalar, i.e., (m Vector a) . Vector b = Vector a . (m Vector b)
  • If Vector a and Vector b are two vectors, then,
    • Vector a . Vector a >= 0
    • |Vector a . Vector b| <= |a| |b|
    • |Vector a + Vector b| <= |a| + |b|
  • If the scalar product of a vector r with each of three non-zero, non-coplanar vectors a, b, and c is zero then r must be a zero vector.

Read More: Coplanar Vector


Things to Remember

  • The product of a scalar and a scalar is always a scalar.
  • The product of a vector and a scalar is always a vector.
  • The product of a vector and a vector is always a scalar.
  • The product of a scalar and a vector is always a vector.
  • The dot product of any of the two unit vectors is equal to the cosine of the angle between their directions.
  • When two vectors have their dot products zero, then they are mutually perpendicular.
  • The dot product of two vectors is commutative.
  • The dot product of two vectors distributes the addition of vectors.

Previous Years Questions

  1. If Vmatrix A A 2 1 A 3 B B 2 1 B 3 C C 2 1 C 3 Vma
  2. For Non Zero Vectors A And B If Left Vec A Vec B R
  3. If The Volume Of The Parallelopiped With A B And C
  4. If A 2 B 3 C 0 Then A B B C C A
  5. If A B A B Then The Angle Between A And B Is
  6. Which Of The Following Expressions Are Meaningfull
  7. U And V Are Unit Vectors Such That U V 0 R Is Any
  8. Three Points 2 1 3 3 5 1 And 1 11 9 Are
  9. There Are N Coplanar Vectors Each Of Magnitude V E
  10. The Volume Of The Tetrahedron Whose Coterminous Ed
  11. The Sine Of The Angle Between The Vectors I 2 J 3
  12. Let U V W Be Such That U 1 V 2 W 3 If The Projecti
  13. Let G Be The Centroid Of A Triangle Abc If Ab A Ac
  14. Let A B C Are Three Non Coplanar Vectors Such That
  15. Let A And B Be Two Unit Vectors If The Vectors C A
  16. If X B C B And X A Then X Is Equal To
  17. If A And B Are Unit Vectors Enclosing An Angle And
  18. If The Difference Of Two Unit Vectors Is Again A U
  19. If Abcdef Is A Regular Hexagon Then Ad Eb Fc Is Eq
  20. For A Non Zero Vector A Which Of The Following Sta

Sample Questions

Ques. Find Vector a .Vector b if Vector a = i + 2j + 3k and Vector b = -2j + 4k. (2 Marks)

Ans. Vector a .Vector b = (i + 2j + 3k) (0i - 2j + 4k) = (1)(0) + (2)(-2) + (3)(4) = - 4 + 12 = 8

Ques. Find Vector a .Vector b if a = (1, 1, 2) and b = (3, 2, -1). (2 Marks)

Ans. a = (1, 1, 2) = i + j + 2k and b = (3, 2, -1) = 3i + 2j -k

Vector a .Vector b = (1)(3) + ((1)(2) + (2)(-1) = 3 + 2 -2 = 3

Ques. Find (Vector a + 3 Vector b) (2 Vector a - Vector b), if Vector a = i + j + 2k and Vector b = 3i + 2j - k. (3 Marks)

Ans. Vector a + 3 Vector b = (i + j + 2k) + 3 (3i + 2j - k) = 10i + 7j - k 2

Vector a - Vector b = 2 (i + j + 2k) - (3i + 2j - k) = - i + 0j + 5k

Therefore, (Vector a + 3 Vector b) (2 Vector a - Vector b) = (10i + 7j - k) (- i + 0j + 5k) = (10)(-1) + (7)(0) + (-1)(5) = -10 + 0 - 5 = -15

Ques. If Vector a and Vector b are two vectors such that |Vector a| = 4, |Vector b| = 3 and Vector a .Vector b = 6, find the angle between the two vectors. (3 Marks)

Ans. Let θ be the angle between Vector a and Vector b.

Then Cos θ = (Vector a .Vector b) / ab = 6 / (4*3) = 1/2

Therefore, cos θ = 1/2 = cos 60 degrees θ = 60 degrees

Ques. Find (Vector a + 2 Vector b) ( Vector a + Vector b), if Vector a = i + j + k and Vector b = 3i + 2j - 2k. (3 Marks)

Ans. (i + j + k) + 2 (3i + 2j - 2k) = 7i + 5j - 3k Vector a + Vector b = (i + j + k) - (3i + 2j - 2k) = - 2i - j + 3k

Therefore, (Vector a + 2 Vector b) (Vector a + Vector b) = (7i + 5j - 3k) (- 2i - j + 3k) = (7)(-2) + (5)(-1) + (-3)(3) = -14 - 5 - 9 = -28

Ques. If Vector a and Vector b are two vectors such that |Vector a| = 5, |Vector b| = 2 and Vector a .Vector b = 10, find the angle between the two vectors. (3 Marks)

Ans. Let θ be the angle between Vector a and Vector b. Then Cos θ = (Vector a .Vector b) / ab = 10 / (5*2) = 1

Therefore, cos θ = 1 = cos 0 degrees θ = 0 degrees

Ques. If Vector a and Vector b are two vectors such that |Vector a| = 0, |Vector b| = 2 and Vector a .Vector b = 6, find the angle between the two vectors. (3 Marks)

Ans. Let θ be the angle between Vector a and Vector b.

Then Cos θ = (Vector a .Vector b) / ab = 6 / (0*2) = infinity

Therefore, cos θ = infinity = cos 90 degrees θ = 90 degrees

Ques. Find the dot product and angle between vectors a and b, given that Vector a = 3i - 2j + k and Vector b = 2i + 3j. (3 Marks)

Ans. Vector a .Vector b = (3i - 2j + k) (2i + 3j + 0k) = (3)(2) - (2)(3) + (1)(0) = 0

Now, Vector a .Vector b = 0,

therefore, the angle between Vector a and Vector b is 90 degrees.

Ques. Find the dot product and angle between vectors a and b, given that Vector a = 2i - 2j - k and Vector b = 6i - 3j + 2k. (3 Marks)

Ans.  Vector a .Vector b = (2i - 2j - k) (6i - 3j + 2k) = (2)(6) - (2)(3) + (-1)(2) = 12 - 6 - 2 = 4

Now Vector a .Vector b = 4

Therefore cos θ = (Vector a .Vector b) / ab Magnitude of Vector a = (22 + 22 + (-1)2)½ = 3 Magnitude of Vector b = (62 + (-3)2 + 22)½ = 7

Therefore, cos θ = 4 / (3*7) So, θ = cos-1 4/21

Ques. Find the angle between vectors a and b, given that Vector a = 2i + 2j and Vector b = 6i + j + 2k. (3 Marks)

Ans. Vector a .Vector b = (2i + 2j + 0k) (6i + j + 2k) = (2)(6) - (2)(1) + (0)(2) = 12 - 2 + 0 = 10

Now Vector a .Vector b = 10

Therefore cos θ = (Vector a .Vector b) / ab Magnitude of Vector a = (22 + 22 + (0)2)½ = 83/2

Magnitude of Vector b = (62 + (1)2 + 22)½ = 411/2

Therefore, cos θ = 4 / (83/2 * 411/2) So, θ = cos-1 4 / (83/2 * 411/2)

Ques. Find Vector a .Vector b if Vector a = 4i + 2j - 3k and Vector b = 3i - 2j + 6k. (2 Marks)

Ans. Vector a .Vector b = (4i + 2j - 3k) (3i - 2j + 6k) = (4)(3) + (2)(-2) + (3)(6) = - 12 - 4 + 18 = 2

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CBSE CLASS XII Related Questions

  • 1.
    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 \(k\).

      • \(0\)
      • \(-2\)
      • \(-1\)
      • \(2\)

    • 2.
      Find:

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

        • \(-\frac{\pi}{2}\)
        • \(-\frac{\pi}{4}\)
        • \(\frac{\pi}{4}\)
        • \(\frac{\pi}{2}\)

      • 3.
        If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


          • 4.
            Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).


              • 5.
                Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


                  • 6.

                    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 \(A'\).

                      • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
                      • \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
                      • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
                      • \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)
                    CBSE CLASS XII Previous Year Papers

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