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Magnitude & Direction Simplified Revision Notes

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11.1.2 Magnitude & Direction

Magnitude and Direction of a Vector

infoNote

In vectors, magnitude and direction are key properties:

  1. Magnitude: The length or size of a vector, denoted a |\mathbf{a}| . For a vector a=(a1a2)\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} in 2D, the magnitude is:
a=a12+a22|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}

In 3D, for a=(a1a2a3)\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}:

a=a12+a22+a32|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}
  1. Direction: The angle the vector makes with a reference axis, often calculated using trigonometry. In 2D, if θ\theta is the angle with the positive xx-axis:
θ=tan1(a2a1)\theta = \tan^{-1}\left(\frac{a_2}{a_1}\right)

Magnitude tells us how long the vector is, while direction indicates where it points. Together, they fully describe the vector's properties.

infoNote

📑Example 1: Find the Direction of the Vector (32)\begin{pmatrix} 3 \\ 2 \end{pmatrix}

  • Diagram Representation: The vector is represented as going from the origin to the point (3, 2).

  • Calculation:

tanα=23\tan \alpha = \frac{2}{3}α=:highlight[33.7°]\alpha = :highlight[33.7°]
infoNote

📑Example 2: Find the Direction of the Vector (42)\begin{pmatrix} -4 \\ -2 \end{pmatrix}

  • Diagram Representation: The vector is represented as going from the origin to the point (4,2).(-4, -2).

  • Calculation:

tanα=24=12\tan \alpha = \frac{2}{4} = \frac{1}{2}α=tan1(12):highlight[26.6°]\alpha = \tan^{-1} \left( \frac{1}{2} \right) \approx :highlight[26.6°]

Since the vector is in the second quadrant, we adjust the angle:

Direction=180°+α180°+26.6°=206.6°:success[207°(3sf)]\text{Direction} = 180° + \alpha \approx 180° + 26.6° = 206.6° \approx :success[207° (3sf)]

Magnitude of a Vector

infoNote

📑Example 1: Find the Magnitude of (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix}

  • Diagram Representation: The vector is represented as going from the origin to the point (2,3)(2, 3).

  • Calculation:

Magnitude=22+32=:success[13]\text{Magnitude} = \sqrt{2^2 + 3^2} = :success[\sqrt{13}]
infoNote

📑Example 2: Find the Magnitude of x\mathbf{x} when x=(24)\mathbf{x} = \begin{pmatrix} -2 \\ -4 \end{pmatrix}

  • Note: This could be worded as "find x|\mathbf{x}|": The bars mean "magnitude".
  • Calculation:
x=(2)2+(4)2=20=:success[25]|\mathbf{x}| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{20} = :success[2\sqrt{5}]

infoNote

📑Example: Write in the form xi+yjxi + yj the vector with direction 120° and magnitude 33.

Diagram Representation:

  • A right triangle is drawn with the angle α\alpha and sides labelled as opposite (OPP), adjacent (ADJ), and hypotenuse (HYP). Notes:
  1. Trigonometric Relations:
  • sinθ=OPPHYP\sin \theta = \frac{\text{OPP}}{\text{HYP}}
  • Therefore, sinθ=OPPl\sin \theta = \frac{\text{OPP}}{l} which implies OPP=lsinθ\text{OPP} = l \sin \theta.
  • cosθ=ADJHYP\cos \theta = \frac{\text{ADJ}}{\text{HYP}}
  • Therefore, cosθ=ADJl\cos \theta = \frac{\text{ADJ}}{l} which implies ADJ=lcosθ\text{ADJ} = l \cos \theta.
  1. Horizontal and Vertical Components:
  • This approach extends to any horizontal and vertical components. Calculations:

  • For the xx-component:

x=3cos120°=:highlight[32]x = 3 \cos 120° = :highlight[-\frac{3}{2}]
  • For the yy-component:
y=3sin120°=:highlight[332]y = 3 \sin 120° = :highlight[\frac{3\sqrt{3}}{2}]
  • Vector Representation:
Vector=:success[32i+332j]\text{Vector} = :success[-\frac{3}{2}i + \frac{3\sqrt{3}}{2}j]

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