Vector components Knowledge-Based Study

Study notes

1. What Are Vector Components?

   In physics, many quantities such as displacement, velocity, and force are vectors, meaning they possess both magnitude and direction. Often, analyzing vectors in two or three dimensions directly can be challenging. To simplify calculations, a two-dimensional vector can be broken down into two independent, perpendicular parts called vector components.

   Typically, we use a standard Cartesian coordinate system with a horizontal x-axis and a vertical y-axis. The projection of a vector along the x-axis is its horizontal component, and its projection along the y-axis is its vertical component. Together, these two perpendicular components are completely equivalent to the original vector.

   Finding Horizontal and Vertical Components

   Consider a vector $\vec{A}$ with magnitude $A$ that makes an angle $\theta$ with the positive x-axis. By forming a right-angled triangle where the vector is the hypotenuse, we can apply basic trigonometry to determine the components:

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

   Note on Angles: Always pay attention to which axis the angle $\theta$ is measured from. If the angle is given relative to the vertical y-axis instead, the trigonometric functions swap roles: the vertical component becomes adjacent ($A_y = A \cos\theta$) and the horizontal component becomes opposite ($A_x = A \sin\theta$).

   Signs of Components: The signs of $A_x$ and $A_y$ depend on the quadrant in which the vector points:

   • Quadrant I: $A_x > 0$, $A_y > 0$
   • Quadrant II: $A_x < 0$, $A_y > 0$
   • Quadrant III: $A_x < 0$, $A_y < 0$
   • Quadrant IV: $A_x > 0$, $A_y < 0$

   Finding Magnitude and Direction

   When the perpendicular components $A_x$ and $A_y$ are known, the original vector $\vec{A}$ can be reconstructed using the Pythagorean theorem and inverse trigonometric functions.

   Magnitude

   The magnitude $A$ of the vector is the length of the hypotenuse:

   $$A = \sqrt{A_x^2 + A_y^2}$$

   Because squares of real numbers are non-negative, the magnitude $A$ is always a positive scalar value (or zero).

   Direction Angle

   The direction angle $\theta$ relative to the x-axis is found using the tangent function:

   $$\tan\theta = \frac{A_y}{A_x} \implies \theta = \arctan\left(\frac{A_y}{A_x}\right)$$

   When using an inverse tangent function, verify the quadrant using the individual signs of $A_x$ and $A_y$ to ensure the angle points in the correct physical direction.

   Sample Problem

   A displacement vector $\vec{d}$ has a magnitude of 50 m and points at an angle of 37° above the positive x-axis. Determine its orthogonal components, and confirm that the reconstructed magnitude equals 50 m.

   Step 1: Calculate the Components

   Using $\cos(37^\circ) \approx 0.80$ and $\sin(37^\circ) \approx 0.60$:

   • $d_x = d \cos(37^\circ) = 50 \times 0.80 = 40\text{ m}$
   • $d_y = d \sin(37^\circ) = 50 \times 0.60 = 30\text{ m}$

   Step 2: Verify by Reconstruction

   Using the Pythagorean theorem:

   $$d = \sqrt{d_x^2 + d_y^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ m}$$

   The reconstructed magnitude matches the initial value, confirming that the components accurately represent the vector.