Vectors — scalar & vector products, unit vector

Page 4: Vectors — Scalar & Vector Products, Unit Vector

Welcome back! In our previous discussions, we learned how to add and subtract vectors. But what about multiplication? It turns out that multiplying vectors isn't as straightforward as multiplying scalars. Vectors have two distinct types of products, each with unique physical significance. Let's dive into the fascinating world of the scalar (dot) product and the vector (cross) product.

{{KEY: type=exam | title=Why Two Types of Multiplication? | text=In physics, we often need to combine two vector quantities to get a result. Sometimes, this result is a scalar (like Work = Force · Displacement), and sometimes it's a new vector (like Torque = Radius × Force). This physical necessity gives rise to two different mathematical operations for multiplying vectors.}}

Scalar vs. Vector Product: The Big Picture

Before we get into the mathematical details, let's start with a high-level comparison. Understanding the fundamental difference between these two products is key to applying them correctly in problems.

The Scalar Product (Dot Product)

The scalar product, or dot product, of two vectors A and B is a scalar quantity obtained by multiplying the magnitude of vector A with the magnitude of vector B and the cosine of the angle θ between them.

{{FORMULA: expr=A · B = |A| |B| cos θ | symbols=|A|: Magnitude of vector A, |B|: Magnitude of vector B, θ: Angle between A and B}}

This formula is incredibly useful for finding the angle between two vectors if you know their components. It's also central to the concept of work in physics, where only the component of force in the direction of displacement does work.

Geometric Interpretation of the Dot Product

Geometrically, the dot product A · B can be visualized in two ways:

1. (Magnitude of A) × (Component of B along A)
2. (Magnitude of B) × (Component of A along B)

The component of B along A is |B| cos θ. This is essentially the "shadow" or projection of vector B onto vector A.

{{VISUAL: diagram: Geometric interpretation of dot product. Vector A is horizontal. Vector B is at an angle θ. A dashed line drops from the tip of B perpendicularly onto A, showing the projection of B onto A, labelled as 'B cos θ'.}}

Properties of the Scalar Product

• Commutative: The order of multiplication does not matter. A · B = B · A
• Distributive: It distributes over vector addition. A · (B + C) = A · B + A · C
• Product with a scalar: (λA) · B = λ(A · B), where λ is a real number.
• Dot product with itself: A · A = |A| |A| cos 0° = |A|². This is a very useful property to find the magnitude of a vector.
• Orthogonal Vectors: If two non-zero vectors A and B are perpendicular (orthogonal), then θ = 90°, and cos 90° = 0. Therefore, A · B = 0. This is a critical test for perpendicularity.
• Parallel Vectors: If two vectors A and B are parallel, then θ = 0°, and cos 0° = 1. Therefore, A · B = |A| |B|.

Dot Product in Component Form

For vectors in Cartesian coordinates, the dot product becomes very simple. Let A = Aₓî + Aᵧĵ + A₂k̂ and B = Bₓî + Bᵧĵ + B₂k̂.

Remember the properties of orthonormal unit vectors î, ĵ, and k̂:

• î · î = ĵ · ĵ = k̂ · k̂ = (1)(1)cos 0° = 1
• î · ĵ = ĵ · k̂ = k̂ · î = (1)(1)cos 90° = 0

Using the distributive property, the dot product simplifies to: A · B = (Aₓî + Aᵧĵ + A₂k̂) · (Bₓî + Bᵧĵ + B₂k̂) A · B = AₓBₓ + AᵧBᵧ + A₂B₂

You simply multiply the corresponding components and add them up!

{{SOLVE: {"problem":"If vector A = 2î + 3ĵ - k̂ and vector B = -î + ĵ + 2k̂, find the angle θ between them.","type":"numerical","subject":"physics","intro":"Yeh ek common NEET question type hai. Chalo, whiteboard pe solve karte hain.","outro":"Simple, hai na? Just formula apply karna hai. Ab class mein wapas chalte hain.","steps":[{"explanation":"First, we need the dot product A · B. We'll multiply the corresponding components and add them.","write":"A · B = (AₓBₓ) + (AᵧBᵧ) + (A₂B₂)"},{"explanation":"Let's substitute the component values from the given vectors.","write":"A · B = (2)(-1) + (3)(1) + (-1)(2) = -2 + 3 - 2 = -1"},{"explanation":"Next, we need the magnitudes of both vectors, |A| and |B|, using the formula |V| = √(Vₓ² + Vᵧ² + V₂²).","write":"|A| = √(2² + 3² + (-1)²) = √(4 + 9 + 1) = √14"},{"explanation":"Now, let's find the magnitude of vector B.","write":"|B| = √((-1)² + 1² + 2²) = √(1 + 1 + 4) = √6"},{"explanation":"Finally, we use the main dot product formula, A · B = |A| |B| cos θ, and rearrange it to solve for cos θ.","write":"cos θ = (A · B) / (|A| |B|)"},{"explanation":"Now, plug in all the values we calculated.","write":"cos θ = -1 / (√14 × √6) = -1 / √84","tough":true,"alt_explanation":"We substitute our results for the dot product and the magnitudes into the angle formula to get our final expression for cos θ."},{"explanation":"To find the angle itself, we take the inverse cosine.","write":"θ = cos⁻¹(-1 / √84)"}]}}}

The Vector Product (Cross Product)

The vector product, or cross product, of two vectors A and B is a third vector, C, whose magnitude is given by the product of the magnitudes of A and B and the sine of the angle θ between them.

The magnitude is: |C| = |A × B| = |A| |B| sin θ

The direction of the resulting vector C is perpendicular to the plane containing vectors A and B. This direction is determined by the Right-Hand Thumb Rule.

{{VISUAL: diagram: Right-Hand Thumb Rule for vector cross product. A right hand is shown with fingers curling from vector A towards vector B through the smaller angle θ. The thumb points upwards, indicating the direction of the resultant vector C = A × B.}}

Right-Hand Thumb Rule

1. Point the fingers of your right hand in the direction of the first vector (A).
2. Curl your fingers towards the direction of the second vector (B) through the smaller angle.
3. Your extended thumb will point in the direction of the resultant vector C = A × B.

{{CALLOUT: type=warning | text=Order Matters! The cross product is anti-commutative. A × B gives a vector pointing in the opposite direction to B × A. So, A × B = - (B × A). Never swap the order unless you also flip the sign!}}

Geometric Interpretation of the Cross Product

The magnitude of the cross product, |A| |B| sin θ, is equal to the area of the parallelogram formed with vectors A and B as adjacent sides.

The area of the triangle formed by vectors A and B as two sides is half of this, i.e., ½ |A × B|.

{{VISUAL: diagram: A parallelogram with adjacent sides represented by vectors A and B. The angle between them is θ. The area is labelled as |A × B|. A diagonal is drawn to show a triangle with area ½ |A × B|. }}

Properties of the Vector Product

• Anti-commutative: A × B = - (B × A)
• Distributive: A × (B + C) = A × B + A × C
• Product with a scalar: (λA) × B = λ(A × B) = A × (λB)
• Cross product with itself: A × A = |A| |A| sin 0° = 0. The cross product of any vector with itself is a null vector.
• Parallel or Anti-parallel Vectors: If two non-zero vectors A and B are parallel (θ=0°) or anti-parallel (θ=180°), then sin θ = 0. Therefore, A × B = 0. This is a critical test for parallelism.
• Orthogonal Vectors: If two vectors are perpendicular (θ=90°), sin 90° = 1. The magnitude of the cross product is maximum: |A × B| = |A| |B|.

Cross Product in Component Form (Determinant Method)

Calculating the cross product using components is most easily done using a determinant. For A = Aₓî + Aᵧĵ + A₂k̂ and B = Bₓî + Bᵧĵ + B₂k̂:

        |  î    ĵ    k̂   |
A × B = | Aₓ   Aᵧ   A₂  |
        | Bₓ   Bᵧ   B₂  |

This expands to: A × B = î(AᵧB₂ - A₂Bᵧ) - ĵ(AₓB₂ - A₂Bₓ) + k̂(AₓBᵧ - AᵧBₓ)

{{VISUAL: diagram: A 3x3 determinant shown for calculating the cross product. Arrows indicate the expansion process for the î, ĵ, and k̂ components. For î, cross out the first row and first column. For ĵ, cross out the first row and second column, and so on.}}

Example 1: Calculating Torque

Given: A force F = (2î + 3ĵ - k̂) N is applied at a point with position vector r = (î - ĵ + 2k̂) m relative to the pivot.

To Find: The torque τ about the pivot.

Approach: Torque is defined as the cross product of the position vector and the force vector, τ = r × F. We will use the determinant method to calculate the cross product.

Working:

1. Set up the determinant:

           |  î    ĵ    k̂   |
   τ =     |  1   -1    2   |
           |  2    3   -1   |
   

2. Expand the determinant: τ = î((-1)(-1) - (2)(3)) - ĵ((1)(-1) - (2)(2)) + k̂((1)(3) - (-1)(2))

3. Simplify each component: τ = î(1 - 6) - ĵ(-1 - 4) + k̂(3 + 2) τ = -5î - ĵ(-5) + 5k̂ τ = -5î + 5ĵ + 5k̂

Final Answer: The torque is τ = (-5î + 5ĵ + 5k̂) N·m.

The Unit Vector

A unit vector is a vector that has a magnitude of exactly 1 and points in a particular direction. It is dimensionless and its sole purpose is to specify a direction in space. It is denoted by a letter with a cap or "hat" on it, like û.

{{KEY: type=definition | title=Unit Vector (â) | text=A dimensionless vector with a magnitude of exactly 1. It is used to specify a direction. The unit vector in the direction of any non-zero vector A is found by dividing the vector by its own magnitude. â = A / |A|.}}

The most common unit vectors are î, ĵ, and k̂, which represent the directions along the positive x, y, and z axes, respectively. Any vector can be expressed as the product of its magnitude and a unit vector in its direction.

A = |A| â

Example 2: Finding a Unit Vector

Given: A vector V = 4î - 3ĵ + 12k̂.

To Find: The unit vector v̂ in the direction of V.

Approach: First, we need to find the magnitude of V, which is |V|. Then, we divide the vector V by its magnitude.

Working:

1. Calculate the magnitude of V: |V| = √(4² + (-3)² + 12²) |V| = √(16 + 9 + 144) |V| = √169 = 13

2. Divide the vector V by its magnitude |V|: v̂ = V / |V| v̂ = (4î - 3ĵ + 12k̂) / 13 v̂ = (4/13)î - (3/13)ĵ + (12/13)k̂

Final Answer: The unit vector is v̂ = (4/13)î - (3/13)ĵ + (12/13)k̂. You can verify that its magnitude is 1.

{{VISUAL: diagram: A vector A in 3D space originating from the origin. A smaller vector, â, is shown along the same line, also originating from the origin, but with its tip on a sphere of radius 1, indicating it is a unit vector.}}

Example 3 (NEET Level): Perpendicular Unit Vector

Given: Two vectors A = 2î + ĵ + k̂ and B = î - ĵ + 2k̂.

To Find: A unit vector which is perpendicular to the plane containing both A and B.

Approach:

1. The cross product A × B gives a vector C that is perpendicular to the plane containing A and B.
2. To get a unit vector in that direction, we just need to find the unit vector of C, which is ĉ = C / |C|.

Working:

1. Calculate the cross product C = A × B:

           |  î    ĵ    k̂   |
   C =     |  2    1    1   |
           |  1   -1    2   |
   

   C = î((1)(2) - (1)(-1)) - ĵ((2)(2) - (1)(1)) + k̂((2)(-1) - (1)(1)) C = î(2 + 1) - ĵ(4 - 1) + k̂(-2 - 1) C = 3î - 3ĵ - 3k̂

2. Now, find the magnitude of C: |C| = √(3² + (-3)² + (-3)²) = √(9 + 9 + 9) = √27 = 3√3

3. Finally, find the unit vector ĉ: ĉ = C / |C| = (3î - 3ĵ - 3k̂) / (3√3) ĉ = (1/√3)î - (1/√3)ĵ - (1/√3)k̂

Final Answer: The required unit vector is (î - ĵ - k̂) / √3. Note that the vector in the opposite direction, (-î + ĵ + k̂) / √3, is also a valid answer.

Common Numerical Traps

Students often make predictable errors when dealing with vector products. Here's a quick guide to avoid them.

NEET UG MCQ Bank

1. If A · B = |A × B|, then the angle between vectors A and B is: a) 0° b) 45° c) 90° d) 180°

💡 Answer: (b) Solution: Given A · B = |A × B|. We know A · B = |A||B| cos θ and |A × B| = |A||B| sin θ. So, |A||B| cos θ = |A||B| sin θ. This simplifies to cos θ = sin θ, or tan θ = 1. The angle θ for which tan θ = 1 is 45° (or π/4 radians).

2. The work done by a force F = (î + 2ĵ + k̂) N in displacing a particle from origin to r = (3î + ĵ - 2k̂) m is: a) 3 J b) 5 J c) 7 J d) 1 J

💡 Answer: (a) Solution: Work done W is the dot product of force F and displacement d. Here d = r - 0 = r. W = F · r W = (î + 2ĵ + k̂) · (3î + ĵ - 2k̂) W = (1)(3) + (2)(1) + (1)(-2) W = 3 + 2 - 2 = 3 J.

3. If vectors P = aî + aĵ + 3k̂ and Q = aî - 2ĵ - k̂ are perpendicular to each other, the positive value of a is: a) 1 b) 2 c) 3 d) 4

💡 Answer: (c) Solution: If two vectors are perpendicular, their dot product is zero. P · Q = 0. (aî + aĵ + 3k̂) · (aî - 2ĵ - k̂) = 0 (a)(a) + (a)(-2) + (3)(-1) = 0 a² - 2a - 3 = 0 This is a quadratic equation. We can factorize it: (a - 3)(a + 1) = 0. The possible values for a are 3 and -1. The question asks for the positive value, which is 3.

4. The area of the parallelogram whose adjacent sides are given by the vectors A = 3î + ĵ + 4k̂ and B = î - ĵ + k̂ is: a) √25 b) √35 c) √42 d) 7

💡 Answer: (c) Solution: The area of the parallelogram is given by the magnitude of the cross product, |A × B|. First, find A × B using the determinant method: A × B = î((1)(1) - (4)(-1)) - ĵ((3)(1) - (4)(1)) + k̂((3)(-1) - (1)(1)) A × B = î(1 + 4) - ĵ(3 - 4) + k̂(-3 - 1) A × B = 5î + ĵ - 4k̂ Now, find the magnitude: |A × B| = √(5² + 1² + (-4)²) = √(25 + 1 + 16) = √42.

Quick Revision Cheatsheet

{{FLASHCARD: q=What is the dot product of 2î and 3ĵ? | a=Zero. The vectors are perpendicular (along x and y axes), so cos 90° = 0. Their dot product (2)(3)cos 90° is 0.}}