Vector Operations and Resolutions

Addition of Vectors

1. Triangle Law of Vector Addition: If two vectors $\vec{A}$ and $\vec{B}$ are represented by the two sides of a triangle taken in order, then their sum $\vec{R}$ is given by the third side in reverse order. $$\vec{R} = \vec{A} + \vec{B}$$

2. Parallelogram Law: If two vectors $\vec{A}$ and $\vec{B}$ act simultaneously at a point, then the resultant is the diagonal of the parallelogram formed: $$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$ where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.

Subtraction of Vectors

To subtract $\vec{B}$ from $\vec{A}$, add $-\vec{B}$ (reverse direction): $$\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$$

Graphical Method: Use triangle or parallelogram by reversing one vector.

Properties of Vector Addition:

• Commutative Law: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$
• Associative Law: $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$
• Existence of Zero Vector: $\vec{A} + \vec{0} = \vec{A}$
• Additive Inverse: $\vec{A} + (-\vec{A}) = \vec{0}$

Example:

Question: Two vectors $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = -\hat{i} + 2\hat{j}$ are added. Find the resultant.

Solution: $$\vec{R} = \vec{A} + \vec{B} = (3 - 1)\hat{i} + (4 + 2)\hat{j} = 2\hat{i} + 6\hat{j}$$ Magnitude: $$|\vec{R}| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$$

Unit Vector

A unit vector is a vector of magnitude 1 and is used to indicate direction only.

Notation: $$\hat{A} = \frac{\vec{A}}{|\vec{A}|}$$ where $\hat{A}$ is the unit vector in the direction of vector $\vec{A}$.

Standard Unit Vectors in 3D:

• $\hat{i}$: Unit vector along x-axis
• $\hat{j}$: Unit vector along y-axis
• $\hat{k}$: Unit vector along z-axis

Any vector can be expressed as: $$\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$$ and its unit vector as: $$\hat{A} = \frac{A_x \hat{i} + A_y \hat{j} + A_z \hat{k}}{\sqrt{A_x^2 + A_y^2 + A_z^2}}$$

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Direction Cosines

Direction cosines are the cosines of the angles made by a vector with the coordinate axes.

If $\alpha$, $\beta$, and $\gamma$ are angles made by a vector $\vec{A}$ with x, y, and z axes respectively: $$\cos\alpha = \frac{A_x}{|\vec{A}|},\quad \cos\beta = \frac{A_y}{|\vec{A}|},\quad \cos\gamma = \frac{A_z}{|\vec{A}|}$$

Property: $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

Example:

Question: A vector has components $\vec{A} = 6\hat{i} - 3\hat{j} + 2\hat{k}$. Find its unit vector and direction cosines.

Solution: Magnitude: $$|\vec{A}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$$

Unit vector: $$\hat{A} = \frac{1}{7}(6\hat{i} - 3\hat{j} + 2\hat{k})$$

Direction cosines: $$\cos\alpha = \frac{6}{7}, \quad \cos\beta = \frac{-3}{7}, \quad \cos\gamma = \frac{2}{7}$$

Resolution of Vectors

Definition: Resolution of a vector means breaking a vector into two or more components that act in different directions and combine to give the original vector.

Rectangular Components

If a vector $\vec{A}$ makes an angle $\theta$ with the x-axis, then it can be resolved into:

• Horizontal component: $A_x = A \cos\theta$
• Vertical component: $A_y = A \sin\theta$

Thus, the vector is expressed as: $$\vec{A} = A_x \hat{i} + A_y \hat{j}$$ $$|\vec{A}| = \sqrt{A_x^2 + A_y^2}, \quad \tan\theta = \frac{A_y}{A_x}$$

Advantages of Vector Resolution:

• Simplifies analysis of motion in 2D.
• Allows easy application of Newton’s laws.
• Makes calculations with forces and fields easier.

Example 1:

Question: A force of 50 N is acting at an angle of 60° above the horizontal. Resolve it into horizontal and vertical components.

Solution: $$F_x = 50 \cos 60^\circ = 50 \cdot 0.5 = 25\ N$$ $$F_y = 50 \sin 60^\circ = 50 \cdot 0.866 = 43.3\ N$$

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Example 2:

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes of 10.0 units. The angle between them is $60^\circ$. Find the magnitude of the resultant vector $\vec{A} + \vec{B}$ and $\vec{A} - \vec{B}$.

Solution for Sum ($\vec{A} + \vec{B}$): Using the Law of Cosines for vector addition: $R^2 = A^2 + B^2 + 2AB\cos(\theta)$ $R^2 = 10^2 + 10^2 + 2(10)(10)\cos(60^\circ)$ $R^2 = 100 + 100 + 200(0.5) = 300$ $R = \sqrt{300} = 10\sqrt{3} \approx 17.32$ units.

Solution for Difference ($\vec{A} - \vec{B}$): The magnitude of the difference is given by: $S^2 = A^2 + B^2 - 2AB\cos(\theta)$ $S^2 = 10^2 + 10^2 - 2(10)(10)\cos(60^\circ)$ $S^2 = 100 + 100 - 200(0.5) = 100$ $S = \sqrt{100} = 10$ units.