Dot Product (Scalar Product)

Dot Product (Scalar Product) - Introduction to Vectors

Welcome, JEE aspirants! The dot product, also known as the scalar product, is a fundamental operation in vector algebra and 3D geometry. Mastering the dot product is crucial for solving a wide range of problems in JEE Main, from finding angles between lines and planes to determining work done by a force. Let's dive in!

Conceptual Explanation and Intuition

The dot product provides a way to multiply two vectors and obtain a scalar (a single number) as the result. Intuitively, it measures the "alignment" or "projection" of one vector onto another. Think of it as how much one vector "contributes" in the direction of the other. If the vectors point in the same direction, the dot product is positive; if they point in opposite directions, it's negative; and if they are perpendicular, it's zero.

Definition: a · b = |a||b|cos θ

The most fundamental definition of the dot product is given by:

where:

• $\vec{a}$ and $\vec{b}$ are the two vectors.
• $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes (lengths) of the vectors.
• $\theta$ is the angle between the two vectors.

This formula directly relates the dot product to the magnitudes of the vectors and the angle between them. If $\vec{a}$ and $\vec{b}$ are unit vectors (magnitude 1), then $\vec{a} \cdot \vec{b}$ is simply $\cos \theta$.

Example: Consider $\vec{a}$ and $\vec{b}$ with magnitudes 5 and 8 respectively, and the angle between them is 60°. Then, $\vec{a} \cdot \vec{b} = 5 \cdot 8 \cdot \cos 60° = 5 \cdot 8 \cdot (1/2) = 20$.

Component Form: a · b = a₁b₁ + a₂b₂ + a₃b₃

In many problems, vectors are given in component form. The dot product can then be calculated as follows:

where:

• $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$
• $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$
• $\hat{i}$, $\hat{j}$, and $\hat{k}$ are the unit vectors along the x, y, and z axes, respectively.

This formula is extremely useful for quickly calculating the dot product when the vectors are given in component form.

Example: If $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - 2\hat{k}$, then $\vec{a} \cdot \vec{b} = (2)(-1) + (-3)(1) + (1)(-2) = -2 - 3 - 2 = -7$.

Angle Between Vectors using Dot Product

Rearranging the definition formula, we can find the angle between two vectors:

This is a powerful tool for finding angles between vectors, lines, and planes. Remember that $\theta$ is always the acute angle between the vectors.

Example: Find the angle between $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$. First, $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) = 0$. Also, $|\vec{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $|\vec{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. Therefore, $\cos \theta = 0 / (\sqrt{2} \cdot \sqrt{2}) = 0$, which means $\theta = 90°$.

Properties of Dot Product

The dot product has several important properties:

1. Commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
2. Distributive: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$
3. Scalar Multiplication: $(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (k\vec{b})$ where $k$ is a scalar.

Important Formulas and Derivations

1. a · a = |a|²

   This follows directly from the definition: $\vec{a} \cdot \vec{a} = |\vec{a}||\vec{a}|\cos 0° = |\vec{a}|^2 \cdot 1 = |\vec{a}|^2$. It is a useful shortcut for calculating the square of the magnitude of a vector.

   $$\vec{a} \cdot \vec{a} = |\vec{a}|^2$$
2. i · i = j · j = k · k = 1

   Since $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors and the angle between each of them and themselves is 0°, $\hat{i} \cdot \hat{i} = |\hat{i}||\hat{i}|\cos 0° = 1 \cdot 1 \cdot 1 = 1$ (and similarly for $\hat{j}$ and $\hat{k}$).

   $$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$$
3. i · j = j · k = k · i = 0

   The unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ are mutually perpendicular. Therefore, the angle between them is 90°. Since $\cos 90° = 0$, we have $\hat{i} \cdot \hat{j} = |\hat{i}||\hat{j}|\cos 90° = 1 \cdot 1 \cdot 0 = 0$ (and similarly for $\hat{j} \cdot \hat{k}$ and $\hat{k} \cdot \hat{i}$).

   $$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$$
4. (a + b) · c = a · c + b · c

   This is the distributive property. This can be easily proved using the component form of the dot product. Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then $\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}$. So, $(\vec{a} + \vec{b}) \cdot \vec{c} = (a_1 + b_1)c_1 + (a_2 + b_2)c_2 + (a_3 + b_3)c_3 = a_1c_1 + b_1c_1 + a_2c_2 + b_2c_2 + a_3c_3 + b_3c_3 = (a_1c_1 + a_2c_2 + a_3c_3) + (b_1c_1 + b_2c_2 + b_3c_3) = \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}$.

   $$(\vec{a} + \vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}$$

Perpendicular (Orthogonal) Vectors: a · b = 0

Two vectors are perpendicular (orthogonal) if and only if their dot product is zero. This is a direct consequence of the definition, since $\cos 90° = 0$.

This is a very important concept for JEE, often used to find conditions for lines and planes to be perpendicular.

Example: Determine if $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ are perpendicular. $\vec{a} \cdot \vec{b} = (2)(1) + (1)(-1) + (-1)(1) = 2 - 1 - 1 = 0$. Therefore, the vectors are perpendicular.

JEE-Specific Tricks

Trick 1: Using Dot Product to Find Projections

The projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by:

The magnitude of the projection is:

Trick 2: Using Dot Product with Cross Product

Sometimes, JEE problems involve a combination of dot products and cross products. Remember that $\vec{a} \cdot (\vec{b} \times \vec{c})$ gives the volume of the parallelepiped formed by the three vectors.

Example (JEE Main level): Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j}$, and $\vec{c} = \hat{i} + 2\hat{j} + \hat{k}$. Find the volume of the parallelepiped formed by these vectors. You would calculate the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$.