Welcome, future engineers and mathematicians! In our journey through Vector Algebra, we've already understood what vectors are, how to add them, and how to multiply them by scalars. Now, we're diving into a crucial concept: the
Scalar Product, often called the
Dot Product. This operation is fundamental in physics and advanced mathematics, giving us a powerful tool to understand the relationship between vectors.
Think of it this way: when you multiply two numbers, you get another number. But what happens when you "multiply" two vectors? There are actually two primary ways – one yields a scalar (a number), and the other yields another vector. Today, we're focusing on the one that gives us a scalar.
Let's begin our deep dive!
### 1. Introduction to Scalar (Dot) Product
The scalar product, as its name suggests, is an operation between two vectors that results in a
scalar quantity (a single real number, with magnitude but no direction). It's denoted by a dot ($cdot$) between the two vectors, hence its alternative name: the
Dot Product.
Why is this useful? Imagine you're pushing a box. The work you do depends not just on how hard you push (force vector) and how far it moves (displacement vector), but also on how much your push is in the direction of motion. If you push sideways, you do no work in the direction of motion! The dot product quantifies this "alignment" or "projection" between two vectors.
### 2. Geometric Definition of Scalar Product
The most intuitive way to understand the dot product is through its geometric definition.
For any two non-zero vectors $mathbf{a}$ and $mathbf{b}$, and let $ heta$ be the angle between them (where $0 le heta le pi$), the scalar product is defined as:
$$ mathbf{a} cdot mathbf{b} = |mathbf{a}| |mathbf{b}| cos heta $$
Let's break down this formula:
* $|mathbf{a}|$ is the
magnitude of vector $mathbf{a}$.
* $|mathbf{b}|$ is the
magnitude of vector $mathbf{b}$.
* $cos heta$ is the
cosine of the angle $ heta$ between the two vectors. The angle $ heta$ is always taken as the smaller angle between the two vectors when their initial points are coincident.
Intuition Check:
* If $ heta = 0^circ$, then $cos heta = 1$. So, $mathbf{a} cdot mathbf{b} = |mathbf{a}| |mathbf{b}|$. This means the vectors are pointing in the exact same direction, and their dot product is the product of their magnitudes. This is the maximum possible value for the dot product.
* If $ heta = 90^circ$ ($pi/2$ radians), then $cos heta = 0$. So, $mathbf{a} cdot mathbf{b} = 0$. This is a crucial property:
two non-zero vectors are perpendicular (orthogonal) if and only if their dot product is zero.
* If $ heta = 180^circ$ ($pi$ radians), then $cos heta = -1$. So, $mathbf{a} cdot mathbf{b} = -|mathbf{a}| |mathbf{b}|$. This means the vectors are parallel but pointing in opposite directions. This is the minimum possible value for the dot product.
The term $|mathbf{b}| cos heta$ represents the
scalar projection of vector $mathbf{b}$ onto vector $mathbf{a}$. Similarly, $|mathbf{a}| cos heta$ is the scalar projection of $mathbf{a}$ onto $mathbf{b}$. So, the dot product can be thought of as the magnitude of one vector multiplied by the scalar projection of the other vector onto it.
Case |
Angle ($ heta$) |
$cos heta$ |
$mathbf{a} cdot mathbf{b}$ |
Interpretation |
|---|
Acute Angle |
$0 < heta < 90^circ$ |
Positive |
Positive |
Vectors generally point in similar directions. |
Right Angle |
$ heta = 90^circ$ |
Zero |
Zero |
Vectors are perpendicular (orthogonal). |
Obtuse Angle |
$90^circ < heta < 180^circ$ |
Negative |
Negative |
Vectors generally point in opposite directions. |
### 3. Algebraic Definition (Component Form) of Scalar Product
While the geometric definition is great for understanding, calculating the dot product using magnitudes and angles can be tedious, especially when vectors are given in component form. This is where the algebraic definition comes in handy.
Let $mathbf{a} = a_1mathbf{i} + a_2mathbf{j} + a_3mathbf{k}$ and $mathbf{b} = b_1mathbf{i} + b_2mathbf{j} + b_3mathbf{k}$ be two vectors in three-dimensional space, where $mathbf{i}, mathbf{j}, mathbf{k}$ are the standard orthonormal basis vectors along the X, Y, and Z axes, respectively.
We know from the geometric definition:
* $mathbf{i} cdot mathbf{i} = |mathbf{i}||mathbf{i}|cos 0^circ = (1)(1)(1) = 1$
* Similarly, $mathbf{j} cdot mathbf{j} = 1$ and $mathbf{k} cdot mathbf{k} = 1$.
* $mathbf{i} cdot mathbf{j} = |mathbf{i}||mathbf{j}|cos 90^circ = (1)(1)(0) = 0$
* Similarly, $mathbf{j} cdot mathbf{k} = 0$ and $mathbf{k} cdot mathbf{i} = 0$.
Now, let's calculate $mathbf{a} cdot mathbf{b}$:
$mathbf{a} cdot mathbf{b} = (a_1mathbf{i} + a_2mathbf{j} + a_3mathbf{k}) cdot (b_1mathbf{i} + b_2mathbf{j} + b_3mathbf{k})$
Using the distributive property (which we'll prove later, but assume for now):
$= a_1mathbf{i} cdot (b_1mathbf{i} + b_2mathbf{j} + b_3mathbf{k}) + a_2mathbf{j} cdot (b_1mathbf{i} + b_2mathbf{j} + b_3mathbf{k}) + a_3mathbf{k} cdot (b_1mathbf{i} + b_2mathbf{j} + b_3mathbf{k})$
$= a_1b_1(mathbf{i} cdot mathbf{i}) + a_1b_2(mathbf{i} cdot mathbf{j}) + a_1b_3(mathbf{i} cdot mathbf{k}) $
$ + a_2b_1(mathbf{j} cdot mathbf{i}) + a_2b_2(mathbf{j} cdot mathbf{j}) + a_2b_3(mathbf{j} cdot mathbf{k}) $
$ + a_3b_1(mathbf{k} cdot mathbf{i}) + a_3b_2(mathbf{k} cdot mathbf{j}) + a_3b_3(mathbf{k} cdot mathbf{k})$
Substitute the values of the dot products of unit vectors:
$= a_1b_1(1) + a_1b_2(0) + a_1b_3(0) $
$ + a_2b_1(0) + a_2b_2(1) + a_2b_3(0) $
$ + a_3b_1(0) + a_3b_2(0) + a_3b_3(1)$
This simplifies to:
$$ mathbf{a} cdot mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 $$
JEE Focus: This algebraic definition is extremely important for practical calculations in coordinate geometry and vector problems. It allows us to compute the dot product directly from the components without needing to find magnitudes and angles explicitly.
### 4. Properties of Scalar Product
The dot product has several important properties that are crucial for solving problems:
1.
Commutativity: The order of vectors does not matter.
$mathbf{a} cdot mathbf{b} = mathbf{b} cdot mathbf{a}$
*Proof (from geometric definition):* $|mathbf{a}||mathbf{b}|cos heta = |mathbf{b}||mathbf{a}|cos heta$.
*Proof (from algebraic definition):* $a_1b_1+a_2b_2+a_3b_3 = b_1a_1+b_2a_2+b_3a_3$.
2.
Distributivity over Vector Addition: The dot product distributes over vector addition.
$mathbf{a} cdot (mathbf{b} + mathbf{c}) = mathbf{a} cdot mathbf{b} + mathbf{a} cdot mathbf{c}$
*This can be proven using the component form, or more elegantly by considering projections.*
3.
Scalar Multiplication: A scalar multiple can be factored out.
$(kmathbf{a}) cdot mathbf{b} = k(mathbf{a} cdot mathbf{b}) = mathbf{a} cdot (kmathbf{b})$, where $k$ is any scalar.
4.
Dot Product of a Vector with Itself:
$mathbf{a} cdot mathbf{a} = |mathbf{a}||mathbf{a}|cos 0^circ = |mathbf{a}|^2$.
*This is an incredibly useful property! It means the square of the magnitude of a vector is equal to its dot product with itself.*
Also, if $mathbf{a} = a_1mathbf{i} + a_2mathbf{j} + a_3mathbf{k}$, then $mathbf{a} cdot mathbf{a} = a_1^2 + a_2^2 + a_3^2 = |mathbf{a}|^2$. This is consistent with the magnitude formula.
5.
Perpendicularity Condition: If $mathbf{a}$ and $mathbf{b}$ are two non-zero vectors, then $mathbf{a} perp mathbf{b}$ (a is perpendicular to b) if and only if $mathbf{a} cdot mathbf{b} = 0$.
Warning: If either $mathbf{a}$ or $mathbf{b}$ is the zero vector, then $mathbf{a} cdot mathbf{b} = 0$, but perpendicularity is not uniquely defined. For the condition to imply perpendicularity, both vectors must be non-zero.
### 5. Applications of Scalar Product
The scalar product is a versatile tool with many applications:
1.
Finding the Angle Between Two Vectors:
From the geometric definition, we can rearrange to find the cosine of the angle:
$$ cos heta = frac{mathbf{a} cdot mathbf{b}}{|mathbf{a}| |mathbf{b}|} $$
This formula allows us to find the angle between any two non-zero vectors using their component form (by calculating $mathbf{a} cdot mathbf{b}$, $|mathbf{a}|$, and $|mathbf{b}|$).
2.
Projection of a Vector onto Another Vector:
This is one of the most significant applications.
*
Scalar Projection (or component of $mathbf{a}$ along $mathbf{b}$): This is the length of the shadow of $mathbf{a}$ on $mathbf{b}$.
$ ext{comp}_{mathbf{b}}mathbf{a} = |mathbf{a}| cos heta = frac{mathbf{a} cdot mathbf{b}}{|mathbf{b}|}$
Note that this is a scalar value and can be negative if the angle is obtuse.
*
Vector Projection (or projection of $mathbf{a}$ onto $mathbf{b}$): This is a vector along $mathbf{b}$ with the magnitude of the scalar projection.
$ ext{proj}_{mathbf{b}}mathbf{a} = left( frac{mathbf{a} cdot mathbf{b}}{|mathbf{b}|}
ight) hat{mathbf{b}} = left( frac{mathbf{a} cdot mathbf{b}}{|mathbf{b}|^2}
ight) mathbf{b}$
Here, $hat{mathbf{b}} = frac{mathbf{b}}{|mathbf{b}|}$ is the unit vector in the direction of $mathbf{b}$.
3.
Work Done by a Constant Force (Physics Application):
If a constant force $mathbf{F}$ causes a displacement $mathbf{d}$, the work done $W$ is given by:
$W = mathbf{F} cdot mathbf{d} = |mathbf{F}| |mathbf{d}| cos heta$
This elegantly captures the idea that only the component of force in the direction of displacement contributes to work.
### 6. Illustrative Examples
Let's solidify our understanding with some examples.
Example 1: Basic Calculation and Angle Finding
Given vectors $mathbf{a} = 2mathbf{i} + 3mathbf{j} - mathbf{k}$ and $mathbf{b} = mathbf{i} - mathbf{j} + 2mathbf{k}$.
Find (a) $mathbf{a} cdot mathbf{b}$ and (b) the angle between $mathbf{a}$ and $mathbf{b}$.
Solution:
(a) Using the component form:
$mathbf{a} cdot mathbf{b} = (2)(1) + (3)(-1) + (-1)(2)$
$= 2 - 3 - 2 = -3$
(b) To find the angle, we need $|mathbf{a}|$, $|mathbf{b}|$, and $mathbf{a} cdot mathbf{b}$.
$|mathbf{a}| = sqrt{2^2 + 3^2 + (-1)^2} = sqrt{4 + 9 + 1} = sqrt{14}$
$|mathbf{b}| = sqrt{1^2 + (-1)^2 + 2^2} = sqrt{1 + 1 + 4} = sqrt{6}$
Now, use the formula $cos heta = frac{mathbf{a} cdot mathbf{b}}{|mathbf{a}| |mathbf{b}|}$:
$cos heta = frac{-3}{sqrt{14} sqrt{6}} = frac{-3}{sqrt{84}} = frac{-3}{2sqrt{21}}$
So, $ heta = cos^{-1}left(frac{-3}{2sqrt{21}}
ight)$. Since $cos heta$ is negative, the angle is obtuse, which makes sense given the negative dot product.
Example 2: Using the Magnitude Property
If $|mathbf{a}|=3$, $|mathbf{b}|=4$, and the angle between $mathbf{a}$ and $mathbf{b}$ is $60^circ$. Find $|mathbf{a}+mathbf{b}|$.
Solution:
We use the property $mathbf{v} cdot mathbf{v} = |mathbf{v}|^2$. So, $|mathbf{a}+mathbf{b}|^2 = (mathbf{a}+mathbf{b}) cdot (mathbf{a}+mathbf{b})$.
Using distributivity:
$(mathbf{a}+mathbf{b}) cdot (mathbf{a}+mathbf{b}) = mathbf{a} cdot mathbf{a} + mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{a} + mathbf{b} cdot mathbf{b}$
Since $mathbf{a} cdot mathbf{b} = mathbf{b} cdot mathbf{a}$ and $mathbf{a} cdot mathbf{a} = |mathbf{a}|^2$, $mathbf{b} cdot mathbf{b} = |mathbf{b}|^2$:
$|mathbf{a}+mathbf{b}|^2 = |mathbf{a}|^2 + |mathbf{b}|^2 + 2(mathbf{a} cdot mathbf{b})$
Now calculate $mathbf{a} cdot mathbf{b}$:
$mathbf{a} cdot mathbf{b} = |mathbf{a}| |mathbf{b}| cos 60^circ = (3)(4)left(frac{1}{2}
ight) = 6$
Substitute the values:
$|mathbf{a}+mathbf{b}|^2 = 3^2 + 4^2 + 2(6)$
$= 9 + 16 + 12 = 37$
Therefore, $|mathbf{a}+mathbf{b}| = sqrt{37}$.
Example 3: Vector Projection
Find the projection of vector $mathbf{a} = 2mathbf{i} + 3mathbf{j} + 2mathbf{k}$ onto vector $mathbf{b} = mathbf{i} + 2mathbf{j} + mathbf{k}$.
Solution:
First, calculate $mathbf{a} cdot mathbf{b}$:
$mathbf{a} cdot mathbf{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10$
Next, find the magnitude of $mathbf{b}$:
$|mathbf{b}| = sqrt{1^2 + 2^2 + 1^2} = sqrt{1 + 4 + 1} = sqrt{6}$
Now, apply the formula for vector projection:
$ ext{proj}_{mathbf{b}}mathbf{a} = left( frac{mathbf{a} cdot mathbf{b}}{|mathbf{b}|^2}
ight) mathbf{b}$
$= left( frac{10}{(sqrt{6})^2}
ight) (mathbf{i} + 2mathbf{j} + mathbf{k})$
$= left( frac{10}{6}
ight) (mathbf{i} + 2mathbf{j} + mathbf{k})$
$= frac{5}{3} (mathbf{i} + 2mathbf{j} + mathbf{k})$
$= frac{5}{3}mathbf{i} + frac{10}{3}mathbf{j} + frac{5}{3}mathbf{k}$
Example 4 (JEE Advanced Type): Conditions for Perpendicularity
If the sum of two unit vectors $mathbf{a}$ and $mathbf{b}$ is a unit vector, prove that the magnitude of their difference is $sqrt{3}$.
Solution:
Given:
1. $mathbf{a}$ is a unit vector $implies |mathbf{a}| = 1 implies |mathbf{a}|^2 = 1$.
2. $mathbf{b}$ is a unit vector $implies |mathbf{b}| = 1 implies |mathbf{b}|^2 = 1$.
3. $mathbf{a} + mathbf{b}$ is a unit vector $implies |mathbf{a} + mathbf{b}| = 1 implies |mathbf{a} + mathbf{b}|^2 = 1$.
We need to find $|mathbf{a} - mathbf{b}|$. Let's start with the given condition $|mathbf{a} + mathbf{b}|^2 = 1$:
$(mathbf{a} + mathbf{b}) cdot (mathbf{a} + mathbf{b}) = 1$
$mathbf{a} cdot mathbf{a} + mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{a} + mathbf{b} cdot mathbf{b} = 1$
$|mathbf{a}|^2 + |mathbf{b}|^2 + 2(mathbf{a} cdot mathbf{b}) = 1$
Substitute $|mathbf{a}|^2 = 1$ and $|mathbf{b}|^2 = 1$:
$1 + 1 + 2(mathbf{a} cdot mathbf{b}) = 1$
$2 + 2(mathbf{a} cdot mathbf{b}) = 1$
$2(mathbf{a} cdot mathbf{b}) = 1 - 2 = -1$
$mathbf{a} cdot mathbf{b} = -frac{1}{2}$
Now, let's find $|mathbf{a} - mathbf{b}|^2$:
$|mathbf{a} - mathbf{b}|^2 = (mathbf{a} - mathbf{b}) cdot (mathbf{a} - mathbf{b})$
$= mathbf{a} cdot mathbf{a} - mathbf{a} cdot mathbf{b} - mathbf{b} cdot mathbf{a} + mathbf{b} cdot mathbf{b}$
$= |mathbf{a}|^2 + |mathbf{b}|^2 - 2(mathbf{a} cdot mathbf{b})$
Substitute $|mathbf{a}|^2 = 1$, $|mathbf{b}|^2 = 1$, and $mathbf{a} cdot mathbf{b} = -frac{1}{2}$:
$|mathbf{a} - mathbf{b}|^2 = 1 + 1 - 2left(-frac{1}{2}
ight)$
$= 2 - (-1) = 2 + 1 = 3$
Therefore, $|mathbf{a} - mathbf{b}| = sqrt{3}$.
JEE Insight: This problem is a classic and demonstrates how the property $mathbf{v} cdot mathbf{v} = |mathbf{v}|^2$ is incredibly powerful for solving problems involving vector magnitudes and sums/differences, especially when angles are not explicitly given but can be derived via the dot product. Also note that $mathbf{a} cdot mathbf{b} = |mathbf{a}||mathbf{b}|cos heta = (1)(1)cos heta = cos heta$. So $cos heta = -1/2$, meaning $ heta = 120^circ$. This geometric insight is helpful but not strictly necessary for the algebraic solution.
### 7. Conclusion
The scalar (dot) product is a cornerstone of vector algebra, linking the geometric properties of vectors (magnitudes and angles) with their algebraic representations (components). Mastering both its geometric definition ($|mathbf{a}||mathbf{b}|cos heta$) and algebraic definition ($a_1b_1 + a_2b_2 + a_3b_3$) is crucial for success in JEE. Its applications range from finding angles and projections to understanding physical concepts like work, making it an indispensable tool in your mathematical toolkit. Keep practicing with diverse problems to build your intuition and problem-solving skills!