In what applications is the dot product commonly used?

The dot product is used in various fields, including physics (work done by a force), geometry (calculating angles and distances), computer graphics (lighting and shading), and engineering (mechanics and structural analysis).

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Dot Product of Two Vectors

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Dot Product

Dot Product, or the inner product or scalar product, is a fundamental operation in vector algebra that combines two vectors to produce a scalar quantity.

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Linear Programming is a powerful mathematical technique used for optimization, where the goal is to find the best outcome in a mathematical model whose requirements are represented by linear relationships.

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Partial derivatives are a key concept in multivariable calculus, essential for understanding how functions change with respect to multiple variables.

Understanding Integrals

Integral is a tool used to add up tiny pieces to find the total area or volume of a continuous shape. The process of calculating an integral...

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Integration is a method to sum functions on a large scale. Here, we discuss integrals of some particular functions commonly used in calculations. These...

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Double Integral

Double integration is a mathematical process used to compute the integral of a function over a two-dimensional region. It involves two successive integrations...

Dot Product of Two Vectors

The dot product, often referred to as the scalar product, is a foundational concept in vector algebra. It combines two vectors to produce a single number (a scalar), offering insights into the relationship between the vectors, such as their relative angle and the projection of one vector onto another.

1.0Dot Product of Two Vectors

The scalar product of two non-zero vectors $a$ and $b$ , denoted by $a$ , $b$ is defined as

$a \cdot b = ∣ a ∥ b ∣ cos θ$

Where, θ is the angle between $a$ and $b$ , 0 $\leq θ \leq π$ .

If either $a = 0$ or $b = 0$ then θ is not defined, and in this case, we define $a \cdot b = 0$

Observations

$a \cdot b$ is a real number
If $a$ and $b$ are non-zero vector, then $a \cdot b = 0$ . If and only if $a$ and $b$ are perpendicular to each other. i.e., $a \cdot b = 0 \Leftrightarrow a ⊥ b$
If θ = 0 then $a \cdot b = ∣ a ∣∣ b ∣$
If θ = π then $a \cdot b = - ∣ a ∣∣ b ∣$
Angle between two non-zero vectors $a$ and $b$ is given by $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ , or $θ = cos^{- 1} (\frac{a \cdot b}{∣ a ∣∣ b ∣})$

2.0Properties of Scalar Products

Since the cosine of 0° is 1 and the cosine of 90° is 0, the scalar products of unit vectors are
$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1.1 cos θ = 1$
$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 1 \cdot 1 cos 9 0^{\circ} = 0$
It also follows commutative and distributive properties
Commutative Property: $a \cdot b = b \cdot a$
$a \cdot (b + c) = a \cdot b + a \cdot c = (b + c) \cdot a$

3.0Dot Product of two Vectors Formula

The dot product can be defined using the magnitudes of the vectors and the cosine of the angle between them. For two vectors $A$ and $B$ , the dot product is given by:

$A \cdot B = ∣ A ∥ B ∣ cos (θ)$

4.0Dot Product of Two Vectors Example

Example 1: Determine the dot product of vectors $A = 2 \hat{i} + 5 \hat{j} - \hat{k}$ and $B = \hat{i} - \hat{j} - 3 \hat{k}$ and the angle between them.

Solution:

$A \cdot B = A_{x} B_{x} + A_{y}$ $B_{y} + A_{z} B_{z}$

$A \cdot B = (2) (1) + (5) (- 1) + (- 1) (- 3)$

= 2 – 5 + 3 = 0

Dot product formula

$A \cdot B = ∣ A ∣∣ B ∣ cos θ$

$∣ A ∣ = 4 + 25 + 1 = 30$

$∣ B ∣ = 1 + 1 + 9 = 11 ∣$

$cos θ = \frac{A \cdot B}{∣ A ∣∣ B ∣} = \frac{0}{30 11}$

cos θ = 0

θ = 90°

So the dot product of two vectors $A & B$ is 0 and the angle between them is 90°.

Example 2: Find $(2 a + 3 b) \cdot (4 a - 6 b)$ , where $a$ , $b$ are mutually perpendicular unit vectors.

Solution: If $a$ and $b$ are mutually perpendicular unit vectors, then $a \cdot b = 0$

So, $(2 a + 3 b) \cdot (4 a - 6 b)$

$= (2 a) (4 a) - (2 a) (6 b) + (3 b) (4 a) - (3 b) (6 b)$

$= 8 (a)^{2} - 12 a \cdot b + 12 b \cdot a - 18 (b)^{2}$

$∵ a \cdot b = 0 = b \cdot a and (a)^{2} = 1 = (b)^{2}$

So, 8 – 18 = –10.

Example 3: If $a = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $b = - \hat{i} + 2 \hat{j} + \hat{k}$ and $c = 3 \hat{i} + \hat{j}$ , then find ‘t’ such that $a + t b$ is perpendicular to $c$ .

Solution:

$a + t b = (2 - t) \hat{i} + (2 + 2 t) \hat{j} + (3 + t) \hat{k}$

Since $a + t b$ is perpendicular to $c$ , so $(a + t b) \cdot c = 0$

∴ (2 – t) × 3 + (2 + 2t) + (3 + t) × 0 = 0

⇒ 6 – 3t + 2 + 2t = 0

⇒ 8 – t = 0 ⇒ t = 0

Example 4: Being given that $∣ a ∣ = 2, ∣ b ∣ = 5$ and angle between $a$ and $b$ is $\frac{2 π}{3}$ . Find the value of a for which the vectors $q = α a + 17 b$ and $q = 3 a - b$ are perpendicular.

Solution: We know the formula of Dot Product formula

$a \cdot b = ∣ a ∣∣ b ∣ cos θ$

$a \cdot b = (2) (5) cos \frac{2 π}{3}$

$= 10 \cdot (- \frac{1}{2}) = - 5$

As given in the question $p$ and $q$ are mutually perpendicular. so $p \cdot q = 0$

So $(α a + 17 b) \cdot (3 a - b) = 0$

$\Rightarrow 3 α (a)^{2} - α a \cdot b + 51 b \cdot a - 17 (b)^{2} = 0$

⇒ 3α(4) – α(–5) + 51(–5) – 17(25) = 0

⇒ 12α + 5α – 255 – 425 = 0

⇒ 17α – 680 = 0

⇒ 17α = 680 ⇒ α = 40.

Example 5: If $∣ a ∣ = 3, ∣ b ∣ = 4$ and $∣ a + b ∣ = 5$ , then find $∣ a - b ∣$

Solution: $∣ a + b ∣ = 5$

Squaring both the sides

$∣ a + b ∣^{2} = 5^{2}$

⇒ $∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b = 25$

⇒ $9 + 16 + 2 a \cdot b = 25$

$\Rightarrow 2 a \cdot b = 0 \Rightarrow a \cdot b = 0$

$a an d b$ are mutually perpendicular

_Now $∣ a - b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b$

_{= 9 + 16 – 0 = 25}

$∣ a - b ∣^{2} = 25$

$∣ a - b ∣ = 5$

Example 6: Find the vector $a$ which is collinear with the vector $b = (3, 6, 6)$ and satisfies the condition $a \cdot b = 27$

Solution: If $a$ is collinear with the $b$ then

$a = (3 α, 6 α, 6 α)$

Now $a \cdot b = 27$

⇒ (3α) (3) + (6α) (6) + (6α) (6) = 27

⇒ 9α + 36α + 36α = 27

⇒ 81α = $27 \Rightarrow α = \frac{27}{81} = \frac{1}{3}$

So $a = (3 \times \frac{1}{3}, 6 \times \frac{1}{3}, 6 \times \frac{1}{3}) = (1, 2, 2)$

5.0Dot Product of Two Vectors Practice Question

Find the angle between the vectors $a - b$ and $a + b$ . If $a = (1, 2, 1) b = (2, - 1, 0)$

Ans: $cos^{- 1} \frac{1}{11}$

Find the angle between the vectors $2 a an d \frac{1}{2} b$ if $a = (- 4, 2, 4)$ and $b = (2, - 2, 0) .$

Ans: $\frac{3 π}{4}$

Find the vector $a$ which is collinear with vector $b = (2, - 1, 0)$ , If $a \cdot b = 10$

Ans: a = (4, –2, 0)

If vector $b$ , which is collinear with the vector $a = (8, - 10, 13)$ forms an acute angle with the unit vector $\hat{k}$ . Being given that $∣ b ∣ = 37$ , find the vector $b$ .

Ans: $b = \frac{8}{3}, \frac{- 10}{3}, \frac{13}{3}$

If the vector $C$ is perpendicular to vectors $a = (2, - 3, 1), b = (1, - 2, 3)$ and satisfies the condition $c \cdot (\hat{i} + 2 \hat{j} - 7 \hat{k}) = 10$ $c .$ then find

Ans: $7 \hat{i} + 5 \hat{j} + \hat{k}$

Find a vector of magnitude $51$ which makes equal angles with the vector $a = \frac{1}{3} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ , $b = \frac{1}{5} (- 4 \hat{i} - 3 \hat{k})$ and $c = \hat{j}$ .

Ans: $\pm (5 \hat{i} - \hat{j} - 5 \hat{k})$

6.0Sample Questions on Dot Product of Two Vectors

Q. What is the dot product of two vectors?

Ans: The scalar product of two non-zero vectors $a$ and $b$ , denoted by $a \cdot b$ is defined as

$a \cdot b = ∣ a ∥∣∣ b ∣ cos θ$ , Where, θ is the angle between $a$ and $b$ , 0 $\leq θ \leq π$ .

Q. How do you calculate the dot product of two vectors?

Ans: To calculate the dot product of two vectors A and B, multiply their corresponding components and sum the results:

$A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$

Q. What does the dot product represent geometrically?

Ans: Geometrically, the dot product measures the projection of one vector onto another, scaled by their magnitudes. It can also determine the angle between two vectors:

Where $θ$ is the angle between vectors A and B.

Q. What is the significance of a dot product of zero?

Ans: A dot product of zero indicates that the vectors A and B are orthogonal (perpendicular) to each other: $A \cdot B = 0$

Q. How is the dot product used to find the angle between two vectors?

Ans: The angle between two vectors A and B can be found using the dot product formula:

$θ = cos^{- 1} (\frac{A \cdot B}{∣ A ∥∣ B ∣})$

Q. What are the properties of the dot product?

Ans: The dot product has several important properties:

Commutative: $A \cdot B = B \cdot A$
Distributive: $A \cdot (B + C) = A \cdot B + A \cdot C$
Scalar Multiplication: $(c A) \cdot B = c (A \cdot B)$

Q. How does the dot product relate to vector projection?

Ans: The dot product is used to compute the projection of one vector onto another. The projection of vector A onto vector B is given by:

$proj_{B} A = \frac{A \cdot B}{∣ B ∣ ^{2}} B$

Join ALLEN!

(Session 2026 - 27)

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Class

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Choose your goal

Preferred Programs

Preferred Mode

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I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Dot Product

Dot Product, or the inner product or scalar product, is a fundamental operation in vector algebra that combines two vectors to produce a scalar quantity.

Linear Programming: A Comprehensive Study

Linear Programming is a powerful mathematical technique used for optimization, where the goal is to find the best outcome in a mathematical model whose requirements are represented by linear relationships.

Partial Derivative

Partial derivatives are a key concept in multivariable calculus, essential for understanding how functions change with respect to multiple variables.

Understanding Integrals

Integral is a tool used to add up tiny pieces to find the total area or volume of a continuous shape. The process of calculating an integral...

Integrals of Particular Functions

Integration is a method to sum functions on a large scale. Here, we discuss integrals of some particular functions commonly used in calculations. These...

Conjugate of Complex Number

Conjugate of a complex number z=a+i b (where a, b are real numbers) is denoted and defined by z=a-ib...

Double Integral

Double integration is a mathematical process used to compute the integral of a function over a two-dimensional region. It involves two successive integrations...

Dot Product of Two Vectors

The dot product, often referred to as the scalar product, is a foundational concept in vector algebra. It combines two vectors to produce a single number (a scalar), offering insights into the relationship between the vectors, such as their relative angle and the projection of one vector onto another.

1.0Dot Product of Two Vectors

The scalar product of two non-zero vectors $a$ and $b$ , denoted by $a$ , $b$ is defined as

$a \cdot b = ∣ a ∥ b ∣ cos θ$

Where, θ is the angle between $a$ and $b$ , 0 $\leq θ \leq π$ .

If either $a = 0$ or $b = 0$ then θ is not defined, and in this case, we define $a \cdot b = 0$

Observations

$a \cdot b$ is a real number
If $a$ and $b$ are non-zero vector, then $a \cdot b = 0$ . If and only if $a$ and $b$ are perpendicular to each other. i.e., $a \cdot b = 0 \Leftrightarrow a ⊥ b$
If θ = 0 then $a \cdot b = ∣ a ∣∣ b ∣$
If θ = π then $a \cdot b = - ∣ a ∣∣ b ∣$
Angle between two non-zero vectors $a$ and $b$ is given by $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ , or $θ = cos^{- 1} (\frac{a \cdot b}{∣ a ∣∣ b ∣})$

2.0Properties of Scalar Products

Since the cosine of 0° is 1 and the cosine of 90° is 0, the scalar products of unit vectors are
$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1.1 cos θ = 1$
$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 1 \cdot 1 cos 9 0^{\circ} = 0$
It also follows commutative and distributive properties
Commutative Property: $a \cdot b = b \cdot a$
$a \cdot (b + c) = a \cdot b + a \cdot c = (b + c) \cdot a$

3.0Dot Product of two Vectors Formula

The dot product can be defined using the magnitudes of the vectors and the cosine of the angle between them. For two vectors $A$ and $B$ , the dot product is given by:

$A \cdot B = ∣ A ∥ B ∣ cos (θ)$

4.0Dot Product of Two Vectors Example

Example 1: Determine the dot product of vectors $A = 2 \hat{i} + 5 \hat{j} - \hat{k}$ and $B = \hat{i} - \hat{j} - 3 \hat{k}$ and the angle between them.

Solution:

$A \cdot B = A_{x} B_{x} + A_{y}$ $B_{y} + A_{z} B_{z}$

$A \cdot B = (2) (1) + (5) (- 1) + (- 1) (- 3)$

= 2 – 5 + 3 = 0

Dot product formula

$A \cdot B = ∣ A ∣∣ B ∣ cos θ$

$∣ A ∣ = 4 + 25 + 1 = 30$

$∣ B ∣ = 1 + 1 + 9 = 11 ∣$

$cos θ = \frac{A \cdot B}{∣ A ∣∣ B ∣} = \frac{0}{30 11}$

cos θ = 0

θ = 90°

So the dot product of two vectors $A & B$ is 0 and the angle between them is 90°.

Example 2: Find $(2 a + 3 b) \cdot (4 a - 6 b)$ , where $a$ , $b$ are mutually perpendicular unit vectors.

Solution: If $a$ and $b$ are mutually perpendicular unit vectors, then $a \cdot b = 0$

So, $(2 a + 3 b) \cdot (4 a - 6 b)$

$= (2 a) (4 a) - (2 a) (6 b) + (3 b) (4 a) - (3 b) (6 b)$

$= 8 (a)^{2} - 12 a \cdot b + 12 b \cdot a - 18 (b)^{2}$

$∵ a \cdot b = 0 = b \cdot a and (a)^{2} = 1 = (b)^{2}$

So, 8 – 18 = –10.

Example 3: If $a = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $b = - \hat{i} + 2 \hat{j} + \hat{k}$ and $c = 3 \hat{i} + \hat{j}$ , then find ‘t’ such that $a + t b$ is perpendicular to $c$ .

Solution:

$a + t b = (2 - t) \hat{i} + (2 + 2 t) \hat{j} + (3 + t) \hat{k}$

Since $a + t b$ is perpendicular to $c$ , so $(a + t b) \cdot c = 0$

∴ (2 – t) × 3 + (2 + 2t) + (3 + t) × 0 = 0

⇒ 6 – 3t + 2 + 2t = 0

⇒ 8 – t = 0 ⇒ t = 0

Example 4: Being given that $∣ a ∣ = 2, ∣ b ∣ = 5$ and angle between $a$ and $b$ is $\frac{2 π}{3}$ . Find the value of a for which the vectors $q = α a + 17 b$ and $q = 3 a - b$ are perpendicular.

Solution: We know the formula of Dot Product formula

$a \cdot b = ∣ a ∣∣ b ∣ cos θ$

$a \cdot b = (2) (5) cos \frac{2 π}{3}$

$= 10 \cdot (- \frac{1}{2}) = - 5$

As given in the question $p$ and $q$ are mutually perpendicular. so $p \cdot q = 0$

So $(α a + 17 b) \cdot (3 a - b) = 0$

$\Rightarrow 3 α (a)^{2} - α a \cdot b + 51 b \cdot a - 17 (b)^{2} = 0$

⇒ 3α(4) – α(–5) + 51(–5) – 17(25) = 0

⇒ 12α + 5α – 255 – 425 = 0

⇒ 17α – 680 = 0

⇒ 17α = 680 ⇒ α = 40.

Example 5: If $∣ a ∣ = 3, ∣ b ∣ = 4$ and $∣ a + b ∣ = 5$ , then find $∣ a - b ∣$

Solution: $∣ a + b ∣ = 5$

Squaring both the sides

$∣ a + b ∣^{2} = 5^{2}$

⇒ $∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b = 25$

⇒ $9 + 16 + 2 a \cdot b = 25$

$\Rightarrow 2 a \cdot b = 0 \Rightarrow a \cdot b = 0$

$a an d b$ are mutually perpendicular

_Now $∣ a - b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b$

_{= 9 + 16 – 0 = 25}

$∣ a - b ∣^{2} = 25$

$∣ a - b ∣ = 5$

Example 6: Find the vector $a$ which is collinear with the vector $b = (3, 6, 6)$ and satisfies the condition $a \cdot b = 27$

Solution: If $a$ is collinear with the $b$ then

$a = (3 α, 6 α, 6 α)$

Now $a \cdot b = 27$

⇒ (3α) (3) + (6α) (6) + (6α) (6) = 27

⇒ 9α + 36α + 36α = 27

⇒ 81α = $27 \Rightarrow α = \frac{27}{81} = \frac{1}{3}$

So $a = (3 \times \frac{1}{3}, 6 \times \frac{1}{3}, 6 \times \frac{1}{3}) = (1, 2, 2)$

5.0Dot Product of Two Vectors Practice Question

Find the angle between the vectors $a - b$ and $a + b$ . If $a = (1, 2, 1) b = (2, - 1, 0)$

Ans: $cos^{- 1} \frac{1}{11}$

Find the angle between the vectors $2 a an d \frac{1}{2} b$ if $a = (- 4, 2, 4)$ and $b = (2, - 2, 0) .$

Ans: $\frac{3 π}{4}$

Find the vector $a$ which is collinear with vector $b = (2, - 1, 0)$ , If $a \cdot b = 10$

Ans: a = (4, –2, 0)

If vector $b$ , which is collinear with the vector $a = (8, - 10, 13)$ forms an acute angle with the unit vector $\hat{k}$ . Being given that $∣ b ∣ = 37$ , find the vector $b$ .

Ans: $b = \frac{8}{3}, \frac{- 10}{3}, \frac{13}{3}$

If the vector $C$ is perpendicular to vectors $a = (2, - 3, 1), b = (1, - 2, 3)$ and satisfies the condition $c \cdot (\hat{i} + 2 \hat{j} - 7 \hat{k}) = 10$ $c .$ then find

Ans: $7 \hat{i} + 5 \hat{j} + \hat{k}$

Find a vector of magnitude $51$ which makes equal angles with the vector $a = \frac{1}{3} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ , $b = \frac{1}{5} (- 4 \hat{i} - 3 \hat{k})$ and $c = \hat{j}$ .

Ans: $\pm (5 \hat{i} - \hat{j} - 5 \hat{k})$

6.0Sample Questions on Dot Product of Two Vectors

Q. What is the dot product of two vectors?

Ans: The scalar product of two non-zero vectors $a$ and $b$ , denoted by $a \cdot b$ is defined as

$a \cdot b = ∣ a ∥∣∣ b ∣ cos θ$ , Where, θ is the angle between $a$ and $b$ , 0 $\leq θ \leq π$ .

Q. How do you calculate the dot product of two vectors?

Ans: To calculate the dot product of two vectors A and B, multiply their corresponding components and sum the results:

$A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$

Q. What does the dot product represent geometrically?

Ans: Geometrically, the dot product measures the projection of one vector onto another, scaled by their magnitudes. It can also determine the angle between two vectors:

Where $θ$ is the angle between vectors A and B.

Q. What is the significance of a dot product of zero?

Ans: A dot product of zero indicates that the vectors A and B are orthogonal (perpendicular) to each other: $A \cdot B = 0$

Q. How is the dot product used to find the angle between two vectors?

Ans: The angle between two vectors A and B can be found using the dot product formula:

$θ = cos^{- 1} (\frac{A \cdot B}{∣ A ∥∣ B ∣})$

Q. What are the properties of the dot product?

Ans: The dot product has several important properties:

Commutative: $A \cdot B = B \cdot A$
Distributive: $A \cdot (B + C) = A \cdot B + A \cdot C$
Scalar Multiplication: $(c A) \cdot B = c (A \cdot B)$

Q. How does the dot product relate to vector projection?

Ans: The dot product is used to compute the projection of one vector onto another. The projection of vector A onto vector B is given by:

$proj_{B} A = \frac{A \cdot B}{∣ B ∣ ^{2}} B$

Frequently Asked Questions