Class 11 Physics Unit 2 Vectors | FBISE Federal Board | Read and Download

Class 11 Physics – Unit 2: Vectors and Equilibrium (FBISE)

Access high-quality, exam-focused notes for Class 11 Physics Unit 2. This unit covers the essential mathematical tools required for mechanics, strictly following the Federal Board (FBISE) requirements.

Key topics include Vector Addition by Rectangular Components, Scalar and Vector Products, Torque, and the Conditions of Equilibrium. We provide detailed derivations, conceptual reasons for "Check Your Retention" questions, and solved numerical problems.

Master the Right-Hand Rule and Equilibrium problems through our step-by-step guides. Stay updated with past paper solutions and important short questions by joining our learning platforms below.

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2.2 Product of Vectors

When two vector quantities are multiplied, the result can be either a scalar quantity or a vector quantity, depending on the nature of the vectors involved.

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2.2.1 Scalar Product or Dot Product

The scalar product is formed when the product of two vectors results in a scalar quantity. It is denoted by a dot (·) between the symbols.

• Definition: The product of the magnitudes of two vectors and the cosine of the angle $\theta$ between them.
• Formula: $\mathbf{A} \cdot \mathbf{B} = AB \cos \theta$
• Geometrical Interpretation: It is the product of the magnitude of one vector and the component of the second vector in the direction of the first.
  • $\mathbf{A} \cdot \mathbf{B} = A(B \cos \theta)$
  • $\mathbf{A} \cdot \mathbf{B} = B(A \cos \theta)$

Properties of Scalar Product

1. Commutative: The order of multiplication does not matter ($\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$).
2. Orthogonal Vectors: The dot product of two perpendicular vectors ($\theta = 90^\circ$) is zero.
   • Unit vectors: $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$.
3. Parallel Vectors: The dot product is maximum when vectors are parallel ($\theta = 0^\circ$), equal to the product of their magnitudes ($AB$).
   • Self-product: $\mathbf{A} \cdot \mathbf{A} = A^2$.
   • Unit vectors: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$.
4. Antiparallel Vectors: The dot product is negative ($-AB$) when vectors are in opposite directions ($\theta = 180^\circ$).
5. Component Form: $\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$.

Examples of Scalar Product

• Work ($W$): $\mathbf{F} \cdot \mathbf{d} = Fd \cos \theta$
• Power ($P$): $\mathbf{F} \cdot \mathbf{v} = Fv \cos \theta$
• Electric Flux ($\Phi$): $\mathbf{E} \cdot \mathbf{A} = EA \cos \theta$

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2.2.2 Vector Product or Cross Product

The vector product is formed when the product of two vectors results in a vector quantity. It is denoted by a cross (×) between symbols.

• Definition: The product of the magnitudes of two vectors and the sine of the angle $\theta$ between them, including a unit vector $\hat{n}$ representing direction.
• Formula: $\mathbf{A} \times \mathbf{B} = AB \sin \theta \hat{n}$
• Direction: The direction of $\hat{n}$ is always perpendicular to the plane containing vectors $\mathbf{A}$ and $\mathbf{B}$, determined by the Right Hand Rule.

The Right Hand Rule

1. Rotate the fingers of your right hand from the first vector toward the second vector through the smaller angle.
2. The erect thumb points in the direction of the product vector.

Properties of Vector Product

1. Anti-commutative: Changing the order reverses the direction ($\mathbf{A} \times \mathbf{B} = -\mathbf{B} \times \mathbf{A}$).
2. Parallel/Antiparallel Vectors: The cross product of parallel ($\theta = 0^\circ$) or antiparallel ($\theta = 180^\circ$) vectors is a null vector (zero).
   • Self-product: $\mathbf{A} \times \mathbf{A} = 0$.
   • Unit vectors: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$.
3. Perpendicular Vectors: The magnitude is maximum ($AB$) when $\theta = 90^\circ$.
   • Unit vectors: $\hat{i} \times \hat{j} = \hat{k}$; $\hat{j} \times \hat{k} = \hat{i}$; $\hat{k} \times \hat{i} = \hat{j}$.
4. Component Form (Determinant): The cross product can be solved using a matrix determinant of the unit vectors and components.
5. Physical Significance: The magnitude $|\mathbf{A} \times \mathbf{B}|$ represents the area of a parallelogram formed by the two vectors as adjacent sides.

Examples of Vector Product

• Torque ($\tau$): $\mathbf{r} \times \mathbf{F} = rF \sin \theta \hat{n}$
• Angular Momentum ($\mathbf{L}$): $\mathbf{r} \times \mathbf{p} = rp \sin \theta \hat{n}$

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Note: For calculations involving rectangular components, use the distributive property for dot products and the determinant method for cross products.

Solutions and Explanations

Based on the principles of vector algebra discussed in the previous notes, here are the correct options for your quiz:

1. The number of perpendicular components of a force (in 2-D) are:
   • Correct Option: B. 2
   • Reasoning: In a two-dimensional plane, a force is resolved into two mutually perpendicular components, typically along the x-axis and y-axis.
2. $\hat{j} \times \hat{i} = \_\_\_\_\_\_$
   • Correct Option: D. $-\hat{k}$
   • Reasoning: According to the anti-commutative property of vector products, since $\hat{i} \times \hat{j} = \hat{k}$, reversing the order results in $\hat{j} \times \hat{i} = -\hat{k}$.
3. A force of 10 N is making an angle of 30° with the horizontal. Its x-component will be:
   • Correct Option: D. 8.7 N
   • Reasoning: The horizontal (x) component is calculated as $F_x = F \cos \theta$.
     $F_x = 10 \cos 30^\circ = 10 \times 0.866 = 8.66$ N, which rounds to 8.7 N.
4. If two forces of magnitude 3 N and 4 N are acting at right angle to each other then their resultant force will be:
   • Correct Option: B. 5 N
   • Reasoning: For perpendicular vectors, the resultant is found using the Pythagorean theorem: $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ N.
5. Angle between two vectors A and B can be easily determined by:
   • Correct Option: A. dot product
   • Reasoning: The dot product formula $\mathbf{A} \cdot \mathbf{B} = AB \cos \theta$ is rearranged to $\cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{AB}$ to find the angle.
6. For which angle the equation $|\mathbf{A} \cdot \mathbf{B}| = |\mathbf{A} \times \mathbf{B}|$ is correct?
   • Correct Option: B. 45°
   • Reasoning: $|\mathbf{A} \cdot \mathbf{B}| = AB \cos \theta$ and $|\mathbf{A} \times \mathbf{B}| = AB \sin \theta$. These magnitudes are equal when $\sin \theta = \cos \theta$, which occurs at 45°.

Vector Physics - Short Question Answers

2.1 If the cross product of two vectors vanishes, what will you say about their orientation?

The magnitude of a cross product is given by $|\vec{A} \times \vec{B}| = AB \sin \theta$. If the product is zero (vanishes), then $\sin \theta = 0$. This means:

• The vectors are parallel ($\theta = 0^\circ$).
• The vectors are anti-parallel ($\theta = 180^\circ$).
• One or both vectors are null vectors.

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2.2 Find the dot product of unit vectors with each other at (a) 0° and (b) 90°.

Using the formula $\vec{A} \cdot \vec{B} = AB \cos \theta$ where $A=1$ and $B=1$:

• (a) At 0°: $1 \cdot 1 \cdot \cos(0^\circ) = 1 \cdot 1 \cdot 1 = \mathbf{1}$
• (b) At 90°: $1 \cdot 1 \cdot \cos(90^\circ) = 1 \cdot 1 \cdot 0 = \mathbf{0}$

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2.3 Show that scalar product obeys commutative property.

The scalar product is defined as $\vec{A} \cdot \vec{B} = AB \cos \theta$.
If we take $\vec{B} \cdot \vec{A}$, it becomes $BA \cos(-\theta)$. Since $\cos(-\theta) = \cos \theta$, then:
$BA \cos \theta = AB \cos \theta$.
Therefore, $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.

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2.4 Solve by using the properties of dot and cross product: (a) $\hat{i} \cdot (\hat{j} \times \hat{k})$ (b) $\hat{j} \times (\hat{j} \times \hat{k})$?

• (a): Since $\hat{j} \times \hat{k} = \hat{i}$, the expression becomes $\hat{i} \cdot \hat{i} = \mathbf{1}$.
• (b): Since $\hat{j} \times \hat{k} = \hat{i}$, the expression becomes $\hat{j} \times \hat{i} = \mathbf{-\hat{k}}$.

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2.5 If both the dot product and the cross product of two vectors are zero. What would you conclude about the individual vectors?

For the dot product to be zero, the vectors must be perpendicular. For the cross product to be zero, they must be parallel. Since two non-zero vectors cannot be both parallel and perpendicular simultaneously, we must conclude that at least one of the vectors is a null (zero) vector.

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2.6 What are rectangular components of a vector? How they can be found?

Rectangular components are the projections of a vector along mutually perpendicular axes (X, Y, and Z axes). They can be found using trigonometry:

• $A_x = A \cos \theta$
• $A_y = A \sin \theta$

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2.7 Give two examples for each of the scalar and vector product.

Scalar Product:

1. Work done ($W = \vec{F} \cdot \vec{d}$)
2. Electric Flux ($\Phi_E = \vec{E} \cdot \vec{A}$)

Vector Product:

1. Torque ($\vec{\tau} = \vec{r} \times \vec{F}$)
2. Angular Momentum ($\vec{L} = \vec{r} \times \vec{p}$)

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2.8 Show that: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.

By multiplying $\vec{A} = (A_x\hat{i} + A_y\hat{j} + A_z\hat{k})$ and $\vec{B} = (B_x\hat{i} + B_y\hat{j} + B_z\hat{k})$, we use the property that $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$.
All cross-terms (like $A_x B_y \hat{i} \cdot \hat{j}$) become zero, leaving only the components along the same axes:
$\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.

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2.9 What units are associated with the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$?

Unit vectors are dimensionless and have no units. Their magnitude is exactly 1, and their only function is to indicate direction.

Comprehensive Questions

2.1 Resolution of a Vector in 2-D

Resolution of a vector is the process of splitting a single vector into two or more vectors (components) such that their combined effect is the same as the original vector. In 2-D, we usually resolve a vector $\vec{A}$ into two perpendicular (rectangular) components along the x and y axes.

• Horizontal Component ($A_x$): $A_x = A \cos \theta$
• Vertical Component ($A_y$): $A_y = A \sin \theta$

2.2 Dot Product (Scalar Product)

The dot product of two vectors $\vec{A}$ and $\vec{B}$ is a scalar quantity defined as the product of the magnitudes of the two vectors and the cosine of the angle $\theta$ between them:

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

Properties:

• It is commutative: $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$
• For parallel vectors ($\theta = 0^\circ$), it is maximum: $AB$
• For perpendicular vectors ($\theta = 90^\circ$), it is zero.

2.3 Cross Product (Vector Product)

The cross product of two vectors results in a vector whose magnitude is the product of their magnitudes and the sine of the angle between them, and whose direction is perpendicular to the plane containing both vectors:

$$\vec{A} \times \vec{B} = (AB \sin \theta) \hat{n}$$

Properties:

• It is anti-commutative: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$
• For parallel vectors ($\theta = 0^\circ$), the cross product is a null vector.

2.4 Geometric Interpretation of Cross Product

The magnitude of the cross product $|\vec{A} \times \vec{B}|$ represents the area of a parallelogram formed by the two vectors $\vec{A}$ and $\vec{B}$ as adjacent sides.

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Numerical Problems

No.	Solution Steps	Result
2.1	Given: $|\vec{A} \times \vec{B}| = \sqrt{3} (\vec{A} \cdot \vec{B})$ $AB \sin \theta = \sqrt{3} AB \cos \theta$ $\tan \theta = \sqrt{3}$	$\theta = 60^\circ$
2.2	Given: $F_x = 20\text{ N}$, $\theta = 30^\circ$. $F_x = F \cos 30^\circ \implies 20 = F (0.866)$	$F = 23.1\text{ N}$
2.3	Angle between vectors $\theta = 80^\circ - 10^\circ = 70^\circ$. Dot: $(5.5)(4.3) \cos 70^\circ = 8.1$ Cross: $(5.5)(4.3) \sin 70^\circ = 22.2$	$8.1\text{ Nm}$ and $22.2\text{ Nm}$
2.4	$\vec{A} \cdot \vec{B} = 6\sqrt{3}$, $|\vec{A} \times \vec{B}| = 6$ $\tan \theta = \frac{|\text{Cross}|}{|\text{Dot}|} = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}}$	$\theta = 30^\circ$
2.5	$F = 200\text{ N}, \theta = 30^\circ$ $F_x = 200 \cos 30^\circ = 173.2\text{ N}$ $F_y = 200 \sin 30^\circ = 100\text{ N}$	$173.2\text{ N}, 100\text{ N}$