Class 11 > Unit # 02: Kinematics > Resolution of vectors

Class 11 Physics • Unit 2 Kinematics

Resolution of Vector and Rectangular Components

Detailed lecture notes featuring visual vector diagrams, formula derivations, and step-by-step rectangular component analysis.

1. Resolution of Vector

Resolution of Vector: The process of splitting or breaking down a single vector into two or more component vectors along specific axes is called the resolution of a vector.

When a vector is resolved into two mutually perpendicular components (at an angle of $90^\circ$ to each other), those components are specifically known as rectangular components.

---

2. Graphical Representation & Construction

Consider a vector $\vec{v}$ represented by line segment $OA$, making an angle $\theta$ with the positive $X$-axis:

1. Setup Axes: Establish an $X$-axis and a $Y$-axis at the tail ($O$) of vector $\vec{v}$.
2. Drop Perpendicular: Draw a perpendicular line from the head ($A$) of the vector to the $X$-axis, intersecting it at point $B$.
3. Form Components:
   • $OB$ represents the horizontal component ($v_x$).
   • $AB$ represents the vertical component ($v_y$).

Figure: Graphical resolution of vector $\vec{v}$ into rectangular components $v_x$ and $v_y$.

Key Insight: By the head-to-tail rule, adding the horizontal and vertical components yields the original vector: $\vec{v} = v_x\hat{i} + v_y\hat{j}$.

---

3. Derivation of Rectangular Components

In right-angled triangle $OAB$, trigonometric ratios are applied to find the components:

• Horizontal Component ($v_x$): Using $\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{OB}{OA} = \frac{v_x}{v}$

  $$v_x = v \cos\theta$$

• Vertical Component ($v_y$): Using $\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{OA} = \frac{v_y}{v}$

  $$v_y = v \sin\theta$$

---

4. Finding Magnitude and Direction

Magnitude of Vector ($v$): Squaring and adding the two component equations ($v_x^2 + v_y^2 = v^2(\cos^2\theta + \sin^2\theta)$):

$$v = \sqrt{v_x^2 + v_y^2}$$

Direction ($\theta$): Dividing the vertical component equation by the horizontal component equation:

$$\tan\theta = \frac{v_y}{v_x} \implies \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$$