Magnitude of a Vector: Definition | Formula | Solved Examples

Vector quantities are physical quantities that have both direction and magnitude, like displacement, velocity, force, etc.

• The direction represents the way in which the vector is pointing.
• The magnitude of a vector represents its length and is always a positive scalar value.
• For any vector \overrightarrow{A}, its magnitude is denoted as |\overrightarrow{A}|.
• For example, if a force of 5i N works on an object, then its magnitude is 5 N, which signifies that the strength of the force applied is 5 N, and ‘i’ in 5i represents that it is applied in the positive x direction.

The magnitude of a vector (sometimes called the length or norm) is a measure of how long the vector is.

Magnitude of a Vector Formula

Depending upon the information given, different formulas can be used to find the magnitude of a vector.

The following image shows the different methods used to find the magnitude of the vector.

1. Magnitude of a vector given its Components

• If the given vector Ā = xi+ yĵ + zk̂, then the magnitude of vector Ā can be calculated using the Pythagorean theorem

|A| = \sqrt{x^2 + y^2 + z^2 }

This formula extends to any number of dimensions, where the magnitude is the square root of the sum of the squares of all its components.

2. Magnitude of a Vector Between Two Points

• If the starting point vector is (x₁, y₁) and the endpoint of a vector is (x₂, y₂)  are given then the magnitude of the vector \overrightarrow{AB} is given by,

The magnitude of a vector, when the start and endpoints of a vector are given, is nothing but the distance between the points. The formula for finding magnitude is given by

\overrightarrow{AB} \ = \ \sqrt{((x_2-x_1)^2+(y_2-y_1)^2)}

For 3D space, if the points are (x₁, y₁, z₁₎ and (x₂, y₂, z₂), the formula becomes:

|\overrightarrow{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2 + (z_2- z_1)^2 }

3. Magnitude of a Position Vector (From Origin)

• If any of the starting or endpoint of a vector is at the origin O(0, 0) and another point is A(x, y), as specified in the figure below,

Then the formula for finding the magnitude of a vector where one of the ends of a vector is at the origin is given by

|\vec{A}| = \sqrt{x^2+y^2}

Similarly, in 3D space, if the endpoint is A(x, y, z), the magnitude is:

|\overrightarrow{A}| = \sqrt{x^2 + y^2 + z^2 }

How to Find the Magnitude of a Vector?

The magnitude of the vector is calculated using the steps discussed below,

Step 1: Identify the x, y, and z components of the vector.
Step 2: Find the square of all the x, y, and z components.
Step 3: Add all the squares found in Step 2.
Step 4: Find the square root of the sum obtained in Step 3. 

The value obtained after step 4 is the magnitude of the given vector.

Example: Find the magnitude of the vector A = 3i + 4j

Solution:

The magnitude of vector A is calculated using the steps discussed above.

Step 1: Comparing A = 3i + 4j with xi + yj we get x = 3 and y = 4
Step 2: x² = 3²  = 9 and y² = 4² = 16
Step 3: x² + y² = 9 + 16 = 25
Step 4: √(25) = 5

Thus, the magnitude of the vector  A = 3i + 4j is 5 units.

➣ Read More,

• Vector Algebra
• Scalar and Vector
• Vector Operations
• Vector Space

Solved Examples on Magnitude of Vector

Example 1: Find the magnitude of the vector Ā = 2i + 3ĵ + 4k.
Solution:

Given, Ā = 2i + 3ĵ + 4k

Magnitude |A| = \sqrt{(2^2+3^2+4^2)}
=\sqrt{(4+9+16)}  
= √29
= 5.38 

The magnitude of vector 2i+ 3ĵ + 4k is 5.38 unit

Example 2: Find the magnitude for the vector Ā = 3i + 3ĵ - 6k.
Solution:

Given, Ā = 3i + 3ĵ - 6k

Magnitude |A| = \sqrt{(3^2+3^2+(-6)^2)}
= \sqrt{(9+9+36)}      
= √54
= 7.35

The magnitude of vector 3i+ 3ĵ - 6k is 7.35 unit.

Example 3: Find the magnitude of the vector if the starting point of a vector is (3, 4) and the ending point is (6, 2).
Solution:

Given,

(x₁, y₁) = (3, 4)
(x₂, y₂) = (6, 2)

|Ā|= \sqrt{((x_2-x_1)^2+(y_2-y_1)^2)}
= \sqrt{((6-3)^2+(2-4)^2)}      
= √(3² + (-2)²)
= √(9+4)
= √13
= 3.6

Thus, the magnitude of the given vector is 3.6 unit.

Example 4: Find the magnitude of the vector if the starting point of a vector is (2, 1, 4) and the ending point is (5, 2, 6).
Solution:

Given,

(x₁, y₁, z₁) = (2, 1, 4)
(x₂, y₂, z₂) = (5, 2, 6)

|Ā| = \sqrt{((x_2-x_1)^2+(y_2-y_1)^2)}
= \sqrt{((5-2)^2+(2-1)^2+(6-4)^2)}      
= \sqrt{(3^2+1^2+2^2)}
= √(9 +1 + 4) 
= √14 = 3.74

Thus, the magnitude of the given vector is 3.74 unit.

Example 5: What is the magnitude of the vector that starts at the origin and endpoint at (3, 4).
Solution:

Given,

Starting Point of vector is O(0, 0)

End Point (x, y) = (3, 4)

Magnitude of Vector (|Ā|) = √(x²+y²)
= √(3² + 4²)
= √(9 + 16)
= √25 = 5

Thus, the magnitude of the given vector is 5 unit.

Example 6: Find the magnitude of the vector in which one of the endpoints is at the origin and the other point is at (1, 4, 3).
Solution:

Given,

End Point of vector is O(0, 0)
Other Point (x, y, z) = (1, 4, 3)

Magnitude of Vector (|Ā|) = √(x²+y²+z²)
= \sqrt{(1^2+4^2+3^2)}      
 = \sqrt{(1+16+9)}      
 = √26 = 5.09

Thus, the magnitude of the given vector is 5.09 unit.

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