Notes : Scalar product or Dot product of two vectors , It's Properties - Param Himalaya

Scalar product or dot product of two vectors :

The scalar product of two vectors $\vec{A}$ and $\vec{B}$ is defined as the product of the magnitudes of $\vec{A}$ and $\vec{B}$ multiplied by the cosine of the smaller angle between them.

It is represented by $\vec{A} \cdot \vec{B}$ and is read as $\vec{A}$ dot $\vec{B}$.

If $\theta$ be the angle between vectors $\vec{A}$ and $\vec{B}$, then

$$\vec{A} \cdot \vec{B} = AB \cos \theta$$

or

$\vec{A} \cdot \vec{B} = A (B \cos \theta)$ 

= magnitude of vector $\vec{A}$ × component of vector $\vec{B}$ in the direction of $ \vec{A}$

Thus, dot product of two vectors is defined as the product of the magnitude of the first vector and the magnitude of the component of second vector in the direction of first vector.

Examples of physical quantities given by the dot product of two vector quantities

(i) The dot product of force ($\vec{F}$) and displacement ($\vec{S}$) gives a quantity known as work W(scalar quantity).

That is,

$$\vec{F} \cdot \vec{S} = W$$

(ii) The dot product of force ($\vec{F}$) and velocity ($\vec{v}$) is equal to power P (scalar quantity).

That is,

$$\vec{F} \cdot \vec{v} = P$$

PROPERTIES OF SCALAR PRODUCT OR DOT PRODUCT OF TWO VECTORS

(i) Dot product is commutative

By definition, $\vec{A} \cdot \vec{B} = AB \cos \theta$ and $\vec{B} \cdot \vec{A} = BA \cos \theta = AB \cos \theta$

$\therefore \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$

which is commutative law

(ii) Dot product is distributive over the addition of vectors

That is, $\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$

(iii) Dot product of two parallel vectors

The angle between two parallel vectors is zero ($\theta = 0^\circ$).

Let vectors $\vec{A}$ and $\vec{B}$ be two parallel vectors, then

$\therefore \vec{A} \cdot \vec{B} = AB \cos 0^\circ = AB$ [$\because \cos 0^\circ = 1$]

Thus, dot product of two parallel vectors is equal to the product of the magnitude of the two vectors.

(iv) Dot product of two equal vectors

The angle between two equal vectors is zero ($\theta = 0^\circ$).

The dot product of two equal vectors is given by

$$\vec{A} \cdot \vec{A} = AA \cos 0^\circ = A^2$$

Dot products of two identical unit vectors

$\hat{i} \cdot \hat{i} = 1 \times 1 \times \cos 0^\circ = 1$ ; $\hat{j} \cdot \hat{j} = 1 \times 1 \times \cos 0^\circ = 1$ ; $\hat{k} \cdot \hat{k} = 1 \times 1 \times \cos 0^\circ = 1$

Thus, $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$

(v) Dot product of perpendicular vectors

The angle between two perpendicular vectors is $90^\circ$ ($\theta = 90^\circ$).

If two vectors $\vec{A}$ and $\vec{B}$ are perpendicular, 

then $\vec{A} \cdot \vec{B} = AB \cos 90^\circ$

 $= 0 ( \because \cos 90^\circ = 0)$

Thus, two vectors are perpendicular if their dot product is zero.

Dot products of two perpendicular unit vectors

$\hat{i} \cdot \hat{j} = 1 \times 1 \times \cos 90^\circ = 0$ ; $\hat{j} \cdot \hat{k} = 1 \times 1 \times \cos 90^\circ = 0$ ; $\hat{k} \cdot \hat{i} = 1 \times 1 \times \cos 90^\circ = 0$

Thus, $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

(vi) Dot product of two anti-parallel vectors

The angle between two anti-parallel vectors is $180^\circ$ ($\theta = 180^\circ$)

If vectors $\vec{A}$ and $\vec{B}$ are anti-parallel vectors, then 

$\vec{A} \cdot \vec{B} = AB \cos 180^\circ$

$= -AB(\because \cos 180^\circ = -1)$

Thus, $(\hat{i}) \cdot (-\hat{i}) = (\hat{j}) \cdot (-\hat{j}) = \hat{k} \cdot (-\hat{k}) = -1$

(vii) Dot product of two vectors in terms of their components

The vectors $\vec{A}$ and $\vec{B}$ in terms of their components are written as

$\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and 

$\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$

$\vec{A}.\vec{B} =A_x\hat{i}+A_y\hat{j}+A_z\hat{k}) \cdot (B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$

$= A_x B_x (\hat{i} \cdot \hat{i}) + A_x B_y (\hat{i} \cdot \hat{j}) + A_x B_z (\hat{i} \cdot \hat{k})$

$+ A_y B_x (\hat{j} \cdot \hat{i}) + A_y B_y (\hat{j} \cdot \hat{j}) + A_y B_z (\hat{j} \cdot \hat{k})$

$+ A_z B_x (\hat{k} \cdot \hat{i}) + A_z B_y (\hat{k} \cdot \hat{j}) + A_z B_z (\hat{k} \cdot \hat{k})$

But $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and 

$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

$\therefore \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$

Numerical Example 1. 

Show that $\vec{A} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} - 2\hat{j} - \hat{k}$ are perpendicular to each other.

SOLUTION. Here, $\vec{A} = 2\hat{i} + \hat{j} + 2\hat{k}$, $\vec{B} = 2\hat{i} - 2\hat{j} - \hat{k}$

Vectors \(\vec{A}\) and \(\vec{B}\) will be perpendicular to each other if \(\vec{A} \cdot \vec{B} = 0\)

Now \(\vec{A} \cdot \vec{B} = (2\hat{i} + \hat{j} + 2\hat{k}) \cdot (2\hat{i} - 2\hat{j} - \hat{k})\)

$= 4(\hat{i} \cdot \hat{i}) - 4(\hat{i} \cdot \hat{j}) - 2(\hat{i} \cdot \hat{k})$

$+ 2(\hat{j} \cdot \hat{i}) - 2(\hat{j} \cdot \hat{j}) - (\hat{j} \cdot \hat{k})$

$+ 4(\hat{k} \cdot \hat{i}) - 4(\hat{k} \cdot \hat{j}) - 2(\hat{k} \cdot \hat{k})$

But \(\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\) and \(\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\)

\(\therefore \vec{A} \cdot \vec{B} = 4 - 2 - 2 = 0\)

Since \(\vec{A} \cdot \vec{B} = 0\), so vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other.

Short Cut Method :

Here, \(\vec{A} = 2\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{B} = 2\hat{i} - 2\hat{j} - \hat{k}\)

\(\therefore A_x = 2, A_y = 1, A_z = 2\) and \(B_x = 2, B_y = -2, B_z = -1\)

Vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other if \(\vec{A} \cdot \vec{B} = 0\)

Now \(\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z\)

= 2 × 2 + 1 × (-2) + 2 × (-1) 

= 4 - 2 - 2 = 0

Thus, \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other.

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Numerical Example 2

. If \(\vec{A} = 2\hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{B} = 3\hat{i} - 2\hat{j} - 4\hat{k}\) , find the dot product of vectors \(\vec{A}\) and \(\vec{B}\).

SOLUTION. \(\vec{A} \cdot \vec{B} = (2\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} - 4\hat{k})\)

$= 6(\hat{i} \cdot \hat{i}) - 4(\hat{i} \cdot \hat{j}) - 8(\hat{i} \cdot \hat{k}) + 6(\hat{j} \cdot \hat{i})$

$- 4(\hat{j} \cdot \hat{j}) - 8(\hat{j} \cdot \hat{k}) + 9(\hat{k} \cdot \hat{i}) - 6(\hat{k} \cdot \hat{j})$

$- 12(\hat{k} \cdot \hat{k})$

Since \((\hat{i} \cdot \hat{i}) = (\hat{j} \cdot \hat{j}) = (\hat{k} \cdot \hat{k}) = 1\) and \((\hat{i} \cdot \hat{j}) = (\hat{j} \cdot \hat{k}) = (\hat{k} \cdot \hat{i}) = 0\)

Hence, eqn. (i) becomes

\(\vec{A} \cdot \vec{B} = 6 - 4 - 12 = - 10\)

Short Cut Method :

Here, \(\vec{A} = 2\hat{i} + 2\hat{j} + 3\hat{k}\), therefore, \(A_x = 2, A_y = 2, A_z = 3\)

\(\vec{B} = 3\hat{i} - 2\hat{j} - 4\hat{k}\), therefore, \(B_x = 3, B_y = -2, B_z = -4\)

$ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$

$= 2 \times 3 + 2 \times (-2) + 3 \times (-4) $

$= 6 - 4 - 12 = -10$