Vectors

What is a Vector?

A vector is a quantity that has both magnitude and direction. Represented typically with an arrow: length shows magnitude and the arrowhead shows direction.

Examples of vectors: displacement, velocity, acceleration, force, momentum.

What is a Scalar Quantity?

A scalar quantity has only magnitude, no direction.

Examples: mass, time, temperature, speed.

How to Represent a Vector

Vectors can be represented:

Graphically – using arrows
With notations – using components (2D for IJSO)

Components of a Vector

If a vector has magnitude \(A\) and angle \(\theta\) from horizontal, then it's components are

Horizontal component (x): \( A_x = A \cos(\theta) \)
Vertical component (y):\(A_y = A \sin(\theta) \)

Expressed with unit vectors: \(A_x \hat{i} + A_y \hat{j}\)

Understanding Unit Vectors

In two-dimensional space, unit vectors are used to express direction along standard coordinate axes. The unit vector in the horizontal (x) direction is denoted as \(\hat{i}\), and in the vertical (y) direction as \(\hat{j}\). These vectors have a magnitude of 1 and point in the positive direction of their respective axes.

When a vector is broken down into components, each component is associated with a unit vector to indicate its direction. So, a vector \( \vec{A} \) with horizontal and vertical components \( A_x \) and \( A_y \), is written as:

\[ \vec{A} = A_x \hat{i} + A_y \hat{j} \]

This form clearly shows the contribution of the vector along each axis and is especially useful in vector addition and resolving forces in physics problems.

Addition of Vectors Using Components

\(R_x = A_x + B_x\)
\(R_y = A_y + B_y\)
\( R = \sqrt{R_x^2 + R_y^2} \)
\(θ = \tan^{-1}(\frac{R_y}{R_x})\)

A vector 𝐀 has a magnitude of 5 units and is directed along the positive x-axis. A vector 𝐁 has a magnitude of 12 units and is directed along the positive y-axis. Find the magnitude and direction of A + B

Vector A: 5 units along x-axis
Vector B: 12 units along y-axis
Resultant: \(R = \sqrt{5^2 + 12^2} = 13\)
Angle: \(θ = \tan^{-1}(\frac{12}{5}) \approx 67.4^{\circ}\)

Determine the x and y components of a displacement whose magnitude is 30.0 m at a 23° angle from the x-axis.

x component: \(A_x = 30 \cos(23^{\circ}) \approx 27.62 \text{m}\)
y component: \(A_y = 30 \sin(23^{\circ}) \approx 11.72 \text{m}\)

A force of 15 N acts in the direction 45° above the positive x-axis, and another force of 10 N acts 30° above the positive x-axis. Find the resultant force and direction.

\(A_x = 15 \cos(45^{\circ}) \approx 10.61 N\)
\(A_y = 15 \sin(45^{\circ}) \approx 10.61 N\)
\(B_x = 10 \cos(30^{\circ}) \approx 8.66 N\)
\(B_y = 10 \sin(30^{\circ}) = 5 N\)
\(R_x = 10.61 + 8.66 = 19.27 N\)
\(R_y = 10.61 + 5 = 15.61 N\)
\(R = \sqrt{19.27^2 + 15.61^2} \approx 24.8 N\)
\(\theta = \tan^{-1}(\frac{15.61}{19.27}) \approx 39.0^{\circ}\)

Vector D = 3i - 4j and vector: E = 4i - j. Find the magnitude of D + E and the magnitude of D - E.

\(D + E = 7\hat{i} - 5\hat{j} \rightarrow |D + E| = 8.60 N\)
\(D - E = -1\hat{i} - 3\hat{j} \rightarrow |D - E| = 3.16 N\)

A student carries a lump of clay from the first floor (ground level) door of a skyscraper (on Verstappen Street) to the elevator, 30 m away. She then takes the elevator to the 11th floor. Finally, she exits the elevator and carries the clay 15 m back toward Verstappen Street. If the distance between two floors is 4.5m, calculate the displacement of the lump of clay.

\(A_x = 4.5 \times 10 = 45m\)
\(A_y = 30 - 15 = 15m\)
\(R = \sqrt{45^2 + 15^2} \approx 47.43m\)

Written by Thenura Dilruk