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 $\overrightarrow{\mathrm{a}}$ and $\vec{b}$ , denoted by $\vec{a}$, $\vec{b}$ is defined as

$\vec{a}\cdot \vec{b}=|\vec{a}\Vert \vec{b}|\cos \theta$

Where, θ is the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$, 0 $\le \theta \le \pi$.

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

Observations

1. $\vec{a}\cdot \vec{b}$ is a real number
2. If $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are non-zero vector, then $\vec{a}\cdot \vec{b}=0$. If and only if $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are perpendicular to each other.  i.e., $\vec{a}\cdot \vec{b}=0\iff \vec{a}\perp \vec{b}$
3. If θ = 0 then $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|$
4. If θ = π then $\vec{a}\cdot \vec{b}=-|\vec{a}||\vec{b}|$
5. Angle between two non-zero vectors  $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is given by $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$ , or $\theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$ 

2.0Properties of Scalar Products

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

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 $\mathbf{A}$ and $\text{ B }$, the dot product is given by:

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

4.0Dot Product of Two Vectors Example

Example 1: Determine 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}}$

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

= 2 – 5 + 3 = 0

Dot product formula 

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

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

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

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

cos θ = 0

θ = 90°

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

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

Solution: If $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ are mutually perpendicular unit vectors, then $\vec{a}\cdot \vec{b}=0$

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

$=(2\vec{a})(4\vec{a})-(2\vec{a})(6\vec{b})+(3\vec{b})(4\vec{a})-(3\vec{b})(6\vec{b})$

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

$\because \vec{a}\cdot \vec{b}=0=\vec{b}\cdot \vec{a}\text{ and }(\vec{a}{)}^2=1=(\vec{b}{)}^2$

So, 8 – 18 = –10.

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

Solution:

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

Since $\vec{a}+t\vec{b}$ is perpendicular to $\overrightarrow{\mathrm{c}}$, so $(\vec{a}+t\vec{b})\cdot \vec{c}=0$

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

⇒ 6 – 3t + 2 + 2t = 0

⇒ 8 – t = 0 ⇒ t = 0

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

Solution: We know the formula of Dot Product formula 

$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$

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

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

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

So $(\alpha \vec{a}+17\vec{b})\cdot (3\vec{a}-\vec{b})=0$

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

⇒ 3α(4) – α(–5) + 51(–5) – 17(25) = 0

⇒ 12α + 5α – 255 – 425 = 0

⇒ 17α – 680 = 0

⇒ 17α = 680 ⇒ α = 40.

Example 5: If $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{a}+\vec{b}|=5$, then find $|\vec{a}-\vec{b}|$

Solution: $|\vec{a}+\vec{b}|=5$

Squaring both the sides

$|\vec{a}+\vec{b}{|}^2=5^2$

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

⇒ $9+16+2\vec{a}\cdot \vec{b}=25$

$\Rightarrow 2\vec{a}\cdot \vec{b}=0\Rightarrow \vec{a}\cdot \vec{b}=0$

$\overrightarrow{\mathrm{a}}and\vec{b}$ are mutually perpendicular

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

_{= 9 + 16 – 0 = 25}

$|\vec{a}-\vec{b}{|}^2=25$

$|\vec{a}-\vec{b}|=5$

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

Solution: If $\overrightarrow{\mathrm{a}}$ is collinear with the $\overrightarrow{\mathrm{b}}$ then

$\vec{a}=(3\alpha ,6\alpha ,6\alpha )$

Now $\vec{a}\cdot \vec{b}=27$

⇒ (3α) (3) + (6α) (6) + (6α) (6) = 27

⇒ 9α + 36α + 36α = 27

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

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

5.0Dot Product of Two Vectors Practice Question

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

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

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

Ans: $\frac{3\pi }{4}$

3. Find the vector $\overrightarrow{\mathrm{a}}$ which is collinear with vector $\vec{b}=(2,-1,0)$, If $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=10$

Ans: a = (4, –2, 0)

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

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

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

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

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

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 $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ , denoted by $\vec{a}\cdot \vec{b}$ is defined as

$\vec{a}\cdot \vec{b}=|\vec{a}\Vert ||\vec{b}|\cos \theta$, Where, θ is the angle between $\vec{a}$ and $\overrightarrow{\mathrm{b}}$, 0 $\le \theta \le \pi$.

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:

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

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 $\theta$ 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: $\mathbf{A}\cdot \mathbf{B}=0$

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:

$\theta ={\cos }^{-1}\left(\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}\Vert |\mathbf{B}\mid }\right)$

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 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:

${\mathrm{proj}}_{\mathbf{B}}\mathbf{A}=\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{B}{|}^2}\mathbf{B}$