Understanding Vectors in Electrocardiography: A Comprehensive Guide to the Heart’s Electrical Axis

Learn how vectors apply to EKG interpretation, calculate the mean electrical axis using leads I, aVF, and II, and understand the clinical significance of axis deviations. This math isn’t needed for day to day reading of EKGs but if you love math like we do, here goes!

One crucial aspect of EKG interpretation is understanding vectors and how they represent the direction and magnitude of electrical impulses in the heart. This guide starts from the basics of vectors and builds up to calculating the heart’s mean electrical axis, ensuring you gain a solid understanding of this essential concept.

What Is a Vector?

A scalar is a quantity that only has magnitude, or size. Examples of scalars include volume, density, speed, energy, mass, and time.

e.g., the EKG strip is running at 25mm/ second or John is running at 5mph.

A vector on the other hand is quantity that has both magnitude and direction. Examples of vectors include velocity, momentum, force, electromagnetic fields, and weight.

e.g., The speed of the EKG paper is 25 mm per second (defining the magnitude) progressing from left to right on the paper (defining the direction). John is running 5mph North bound is another example.

A vector is typically represented by an arrow:

• Length of the Arrow: Indicates the magnitude.
• Direction of the Arrow: Indicates the direction of the quantity.

Vectors in the Context of an EKG

In an EKG, vectors represent the electrical activity of the heart as it depolarizes and repolarizes:

• Depolarization Vector: Shows the direction and magnitude of electrical impulses as heart muscle cells become electrically active. By convention, a wave of depolarization is considered to be in the direction of the positive electrode.
• Repolarization Vector: Represents the return of heart muscle cells to their resting electrical state. By convention, a wave of repolarization is considered to be in the opposite direction of the positive electrode.

The heart’s electrical activity can be visualized as a series of vectors (since the activity simultaneously spreads in multiple directions) that combine to form the mean electrical axis, reflecting the average direction of electrical depolarization during ventricular contraction. To simplify this, let us consider the following example: One wave of depolarization is left bound by 7mV (so -7mV in lead I) and another is right bound by 12mV (so +12mV in lead I). A electrode in the middle of these two forces would simply record 12 + (-7)= 5mV at an axis of 0 degrees (which is along lead I).

Let us complicate this a lot!

In the image, notice how lead I has a 7 mm positive deflection and 1 mm negative deflection. So the net amplitude is 6 mm positive deflection.

Lead II on the other hand is at 60 degrees to lead I and has a 4 mm positive deflection and a 3 mm negative deflection. So the net amplitude is 1 mm positive deflection.

So we now have two forces: 6 mm at 0 degrees and 1 mm at 60 degrees. So, how do we estimate the direction of depolarization and its amplitude?

Step 1: Understand the Problem

• Net QRS amplitude in Lead I (0°): +6 mm (positive amplitude indicates the depolarization wave is directed toward the positive electrode of Lead I).
• Net QRS amplitude in Lead II (60°): +1 mm (positive amplitude indicates the depolarization wave is directed toward the positive electrode of Lead II).

The goal is to calculate:

1. Magnitude of the resultant depolarization vector (total QRS vector amplitude).
2. Direction (angle) of the resultant depolarization vector relative to Lead I (0°).

Step 2: Represent the Leads as Vectors

Each lead measures the projection of the heart’s electrical depolarization vector onto its axis. To find the resultant vector, we must resolve these projections into their x (horizontal) and y (vertical) components and then combine them.

Lead I (0 degrees)

• Lead I aligns with the x-axis (horizontal).
• The x-component of the depolarization vector is entirely determined by the amplitude in Lead I: XLead I=+6 mm.
• The y-component of the vector in Lead I is zero: YLead I=0 mm.

Lead II (60 degrees)

• The amplitude in Lead II is a combination of both x- and y-components of the depolarization vector. Using trigonometry:
• The x-component is: XLead II=AmplitudeLead II × cos⁡(60∘)=1 mm×0.5= 0.5 mm. How you ask?
  • Note that we know the hypotenuse and need to find out the adjacent side. The angle between the two is 60 degrees.
  • Recollect from math days that cos (short for cosine)= adjacent side/ hypotenuse. Cos 60 is 0.5
  • In this case, cos 60 (since the lead is at a 60 degree angle)= adjacent side (the x component)/ 1mm (the total amplitude on lead 2).
  • The value of Cos 60 is 0.5. So XLeadII/1= 0.5 or XLeadII is 0.5mm
• The y-component is: YLead II=AmplitudeLead II × cos⁡(30)=1 mm×0.866=0.866 mm. How you ask?
  • Note that we know the hypotenuse and need to find out the adjacent side. The angle between the two is 30 degrees.
  • Cos 30 = 0.866
  • In this case, cos 30 (since the lead is at a 30 degree angle to Y-axis)= adjacent side (the Y component)/ 1mm (the total amplitude on lead 2).
  • The value of Cos 30 is 0.866. So YLeadII/1= 0.866 or YLeadII is 0.866 mm

Step 3: Combine the Components

The total x-component of the resultant depolarization vector is the sum of the x-components from Lead I and Lead II:

Xtotalv=vXLead I + XLead II = 6+0.5 =6.5 mm.

The total y-component of the resultant depolarization vector is the y-component from Lead II (since Lead I contributes no y-component):

Ytotal=YLead II=0.866 mm.

Step 4: Calculate the Resultant Vector’s Magnitude

Since we have the X and Y axes, we can use the Pythagorean theorem:

(Hypotenuse)^2= (Side 1^2)+ (Side 2^2)

Hypotenuse= ((side 1^2)+(side 2^2))^(1/2)= ((6.5^2)+(0.866^2))^(1/2)= 6.56 mm

Step 5: Calculate the Direction (Angle)

Finally, we need to estimate the direction of the QRS vector: We can estimate this relatively easily since we already know the opposite side and the adjacent side of the vector.

If you recollect, cot(θ) = adjacent side/ opposite side. In this case, θ is the angle to the QRS vector to lead I. So the opposite side is amplitude along the Y-axis and the adjacent side is the amplitude along the X axis.

So, cot(θ)= 6.5/0.866=7.506

So θ= about 7.6 degrees.

Combining steps 4 and 5, we now know that the axis of the EKG is about 7.6 degrees and the amplitude is about 6.56 mm. Since 1mm is 0.1 mV, the voltage is 0.656 mV.

If you are wondering where you get the numbers for cos, cot, etc. just google them!

Finally, if we were to use an augmented limb lead in the calculation, the voltage will need to be adjusted since the augmented output is typically boosted by 50%. So, if we notice a 3mm net amplitude on aVF, then the actual amplitude is 3/(1+50%)= 2 mm along the Y axis and 0 along the X-axis.

The good news is that if you’ve stuck around thus far, you now understand exactly what the axis means. The great news is that you rarely if ever need measurements this precise! As we go on, we will review how to intuitively get a good enough estimate for the axes!