Dot Product

The 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. This operation is essential in various fields, including physics, engineering, and computer graphics, due to its ability to measure the similarity between two vectors. The dot product can provide information about the angle between vectors, the projection of one vector onto another, and more.

1.0Dot and Cross Product of a Vector

What is the Dot Product?

The Dot Product $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$ is calculated as the sum of the products of corresponding components of two vectors.

The dot product is also known as the inner product or the scalar product and reads “a dot b”.

If $\overrightarrow{\mathrm{a}}={\mathrm{x}}_1\hat{\mathrm{i}}+{\mathrm{y}}_1\hat{\mathrm{j}}+{\mathrm{z}}_1\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}={\mathrm{x}}_2\hat{\mathrm{i}}+{\mathrm{y}}_2\hat{\mathrm{j}}+{\mathrm{z}}_1\hat{\mathrm{k}}$ then dot product is $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}={\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{y}}_1{\mathrm{y}}_1+{\mathrm{z}}_1{\mathrm{z}}_2.$

The scalar product is useful for determining the angle between two vectors.

Geometric Interpretation

Geometrically, the dot product can be interpreted as:

$\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}\Vert \mathbf{B}|\cos (\theta )$

Where $|\mathbf{A}|and|\mathbf{B}|$ are the magnitudes of the vectors, and $\theta$ is the angle between them.

What is the Cross Product?

Given two vectors $\mathbf{A}=\left({\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3\right)$ and $\mathbf{B}=\left({\mathrm{B}}_1,{\mathrm{B}}_2,{\mathrm{B}}_3\right)$, the cross product $\mathbf{A}\times \mathbf{B}$ is defined as:

$\mathbf{A}\times \mathbf{B}=\left({\mathrm{A}}_2{\mathrm{B}}_3-{\mathrm{A}}_3{\mathrm{B}}_2,{\mathrm{A}}_3{\mathrm{B}}_1-{\mathrm{A}}_1{\mathrm{B}}_3,{\mathrm{A}}_1{\mathrm{B}}_2-{\mathrm{A}}_2{\mathrm{B}}_1\right)$

This resultant vector is orthogonal to both $\mathbf{A}and\text{ B }$.

Geometric Interpretation

The magnitude of the cross product is given by: $\mathbf{A}\times \mathbf{B}|=|\mathbf{A}\Vert \mathbf{B}\mid \sin (\theta )$ where $|\mathbf{A}|$ and $|\mathbf{B}|$are the magnitudes of vectors $\mathbf{A}$ and $\text{ B }$, and θ is the angle between them determines the direction of the resulting vector, following the right-hand rule.

2.0Dot Product Formula

For two vectors $\mathbf{A}=\left({\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3\right)$ and $\mathbf{B}=\left({\mathrm{B}}_1,{\mathrm{B}}_2,{\mathrm{B}}_3\right)$ in three-dimensional space, the dot product \mathbf{A} \cdot \mathbf{B}is defined as:

$\mathbf{A}\cdot \mathbf{B}={\mathrm{A}}_1{\mathrm{B}}_1+{\mathrm{A}}_2{\mathrm{B}}_2+{\mathrm{A}}_3{\mathrm{B}}_3$

3.0Dot Product Rules

The dot product, commonly known as the scalar product, is a foundational operation in vector algebra with several important properties and rules. Here are the key rules and properties of the dot product:

1. Commutative Property

The dot product of 2 vectors is commutative, meaning the order in which you take the dot product does not matter: 

$\mathbf{A}\cdot \mathbf{B}=\mathbf{B}\cdot \mathbf{A}$

2. Distributive Property

The dot product distributes over vector addition: 

$\mathbf{A}\cdot (\mathbf{B}+\mathbf{C})=\mathbf{A}\cdot \mathbf{B}+\mathbf{A}\cdot \mathbf{C}$

3. Scalar Multiplication

If you multiply a vector by a scalar and then take the dot product, it is equivalent to taking the dot product first and then multiplying by the scalar:

$(\mathrm{k}\mathbf{A})\cdot \mathbf{B}=\mathrm{k}(\mathbf{A}\cdot \mathbf{B})\mathbf{A}\cdot (\mathrm{k}\mathbf{B})=\mathrm{k}(\mathbf{A}\cdot \mathbf{B})$

4. Zero Vector

The dot product of any vector with the zero vector results in zero: 

$\mathbf{A}\cdot 0=0$

5. Orthogonal Vectors

If two vectors are orthogonal (perpendicular), their dot product is zero: 

$\mathbf{A}\cdot \mathbf{B}=0\text{ if }\mathbf{A}\perp \mathbf{B}$

6. Magnitude Relation

The dot product of a vector with itself gives the square of its magnitude: $\mathbf{A}\cdot \mathbf{A}=|\mathbf{A}{|}^2$ where $|\mathbf{A}|$ is the magnitude of $\mathbf{A}$.

7. Angle Between Vectors

The dot product can be used to find the cosine of the angle θ between two vectors:

$\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos (\theta )$ This relation can be rearranged to find the angle: $\cos (\theta )=\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}\Vert \mathbf{B}|}$

4.0Dot Product of a Vector with Itself

For a vector $\mathbf{A}=\left({\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3\right)$ in three-dimensional space, the dot product of the vector with itself is given by: $\mathbf{A}\cdot \mathbf{A}={\mathrm{A}}_1^2+{\mathrm{A}}_2^2+{\mathrm{A}}_3^2$

5.0Dot Product of Parallel Vectors

When the two vectors are parallel then 

1. Same Direction:

When vectors $\mathbf{A}and\text{ B }$ are parallel and in the same direction, $\theta =0^{\circ }$. Therefore, $\cos \left(0^{\circ }\right)=1:\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}\Vert \mathbf{B}|$

2. Opposite Direction:

When vectors A and B are parallel but in opposite directions, θ=180.  Therefore,

$\cos \left(180^{\circ }\right)=-1:\mathbf{A}\cdot \mathbf{B}=-|\mathbf{A}\Vert \mathbf{B}|$

6.0Dot Product of Perpendicular Vectors

If 2 vectors are Perpendicular, then their dot product is zero: 

$A\cdot B=0$

Geometric Interpretation

Perpendicular vectors form a 90^∘ angle between them. In terms of the dot product, this means:

$\cos \left(90^{\circ }\right)=0$

Thus, $\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}\Vert \mathbf{B}|\cos \left(90^{\circ }\right)=0.$

7.0Dot Product of Unit Vectors

A unit vector $\hat{\mathbf{u}}$ is a vector with a magnitude of 1. For example, in three-dimensional space, the unit vectors are $\hat{\mathbf{i}}=(1,0,0),\hat{\mathbf{j}}=(0,1,0),and\hat{\mathbf{k}}=(0,0,1).$

1. Same Unit Vector

The dot product of a unit vector with itself is always 1: $\hat{\mathbf{u}}\cdot \hat{\mathbf{u}}=1$

This is because the magnitude of a unit vector is 1. i.e. $|\hat{u}|=1$

2. Different Unit Vectors

The dot product of 2 different unit vectors depends on their orientation relative to each other: $\hat{\mathbf{u}}\cdot \hat{\mathbf{v}}=\cos (\theta )$, where θ is the angle between the two-unit vectors.

• If $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are perpendicular $(\theta =90^{\circ })$, then: $\hat{\mathbf{u}}\cdot \hat{\mathbf{v}}=0$. 
• If $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are parallel $(\theta =0^{\circ }\text{ or }180^{\circ })$, then: $\hat{\mathbf{u}}\cdot \hat{\mathbf{v}}=1$

8.0Dot Product Example

Example 1: Find the dot product of 2 vectors $\overrightarrow{\mathrm{A}}=[3,-2,5],\overrightarrow{\mathrm{B}}=[4,0,-1]$

Solution:

$A\cdot B=A_x{B}_x+A_y{B}_y+A_z{B}_z$

$A\cdot B=(3)(4)+(-2)(0)+(5)(-1)$

$A\cdot B=12+0-5=7$

So, the dot product of A × B is 7.

Example 2: Let there be two vectors |a| = 10 and |b| = 5 and θ = 45°. Find their dot product.

Solution: As we know the formula of dot product

a × b = |a| |b| cos θ

a × b = (10) (5) cos 45°

$=50\times \frac{1}{\sqrt{2}}=25\sqrt{2}$

Example 3: Given vector A = [1, 2, 3] and B = [4, –5, 6], Find the dot product of these two vectors.

Solution: As we know the formula of dot product 

a × b = |a| |b| cos θ

a × b = (1) (4) + (2) (–5) + (3) (6)

= 4 – 10 + 18 = 12

$|A|=\sqrt{1+4+9}=\sqrt{14}$

$|B|=\sqrt{16+25+36}=\sqrt{77}$

$\therefore \cos \theta =\frac{\mathrm{a}\cdot \mathrm{b}}{|\mathrm{a}||\mathrm{b}|}=\frac{12}{\sqrt{14}\sqrt{77}}\approx \frac{12}{32.82}\approx 0.3656$

θ = cos^–1(0.3656) » 68.56°.

Example 4: Calculate the dot product of the two vectors and angle between them A=(-1,2,-2) \quad B=(6,3,-6)

Solution: Dot product

$A\cdot B=A_x{B}_x+A_y{B}_y+A_z{B}_z$

$A\cdot B=(-1)(6)+(2)(3)+(-2)(-6)$

= –6 + 6 + 12 = 12

By dot product formula

$\mathrm{A}\cdot \mathrm{B}=|\mathrm{A}||\mathrm{B}|\cos \theta$

$|\mathrm{A}|=\sqrt{1+4+4}=\sqrt{9}=3$

$|B|=\sqrt{36+9+36}=\sqrt{81}=9$

$\therefore \cos \theta =\frac{\mathrm{A}\cdot \mathrm{B}}{|\mathrm{A}||\mathrm{A}|}$

$\cos \theta =\frac{12}{3\times 9}=\frac{4}{9}$

$\theta ={\cos }^{-1}\left(\frac{4}{9}\right)$.

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

Solution: $\mathrm{A}\cdot \mathrm{B}={\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{y}}+{\mathrm{A}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{z}}$

A × B = (2) (1) + (5) (–1) + (–1) (–3)

= 2 – 5 + 3 = 0

Dot product formula 

$\mathrm{A}\cdot \mathrm{B}=|\mathrm{A}||\mathrm{B}|\cos \theta$

$|A|=\sqrt{4+25+1}=\sqrt{30}$

$|B|=\sqrt{1+1+9}=\sqrt{11}$

$\cos \theta =\frac{A\cdot B}{|A||B|}=\frac{0}{\sqrt{30}\sqrt{11}}$

os θ = 0

θ = 90°

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

9.0Dot Product Practice Questions

Find the dot product of the following vectors and determine the angle between them.

1. A = (2, 4, 1) and B = (3, 5, 7)
2. A = (–2, 3, 11) and B = (5, 7, –4)
3. $A=2\hat{i}+5\hat{j}-\hat{k}andB=\hat{i}-2\hat{j}+\hat{k}$

10.0Solved Questions on Dot Product

Q. What is the dot product?

Ans: The dot product is also known as the inner product or the scalar product and reads “a dot b”.

If $\overrightarrow{\mathrm{a}}={\mathrm{x}}_1\hat{\mathrm{i}}+{\mathrm{y}}_1\hat{\mathrm{j}}+{\mathrm{z}}_1\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}={\mathrm{x}}_2\hat{\mathrm{i}}+{\mathrm{y}}_2\hat{\mathrm{j}}+{\mathrm{z}}_1\hat{\mathrm{k}}$ then dot product is $\vec{a}\cdot \vec{b}=x_1{x}_2+y_1{y}_1+z_1{z}_2.$

Q. How do you calculate the dot product of 2 vectors?

Ans: To calculate the dot product of two vectors, multiply the corresponding components of the vectors and then sum the products. For vectors $\mathbf{A}=\left({\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3\right)$ and $\mathbf{B}=\left({\mathrm{B}}_1,{\mathrm{B}}_2,{\mathrm{B}}_3\right)$:

$\mathbf{A}\cdot \mathbf{B}={\mathrm{A}}_1{\mathrm{B}}_1+{\mathrm{A}}_2{\mathrm{B}}_2+{\mathrm{A}}_3{\mathrm{B}}_3$

Q. How is the dot product geometrically interpreted?

Ans: Geometrically, the dot product of two vectors $\mathbf{A}$ and $\text{ B }$ can be interpreted as: $\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}\Vert \mathbf{B}|\cos (\theta )$ where $|\mathbf{A}|and|\mathbf{B}|$ are the magnitudes of the vectors, and is the angle between them.

Q. What does it mean if the dot product is zero?

Ans: If the dot product of 2 vectors is zero, it means that the vectors are orthogonal (perpendicular) to each other. This is because the cosine of $90^{\circ }(or\frac{\pi }{2}radians)$ is zero: $\mathbf{A}\cdot \mathbf{B}=0\Rightarrow \theta =90^{\circ }$

Q. What are the properties of the dot product?

Ans: The dot product has several important properties:

Commutative: $\mathbf{A}\cdot \mathbf{B}=\mathbf{B}\cdot \mathbf{A}$

Distributive: $\mathbf{A}\cdot (\mathbf{B}+\mathbf{C})=\mathbf{A}\cdot \mathbf{B}+\mathbf{A}\cdot \mathbf{C}$

Scalar Multiplication: $(c\mathbf{A})\cdot \mathbf{B}=c(\mathbf{A}\cdot \mathbf{B})$

Q. How does the dot product relate to the projection of one vector onto another?

Ans: The projection of vector $\mathbf{A}$ onto vector $\text{ B }$ is given by:

${\mathrm{proj}}_{\mathbf{B}}\mathbf{A}=\left(\frac{\mathbf{A}\cdot \mathbf{B}}{\mathbf{B}\cdot \mathbf{B}}\right)\mathbf{B}$

The dot product $\mathbf{A}\cdot \mathbf{B}$ is crucial in determining the magnitude of this projection.