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 Formulas
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) = 5Thus, the magnitude of the vector A = 3i + 4j is 5 units.
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
= 5.38The 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
= 7.35The magnitude of vector 3i+ 3ĵ - 6k is 7.35 unit.
Example 3: 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 = 5Thus, the magnitude of the given vector is 5 unit.
Example 4: 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 = 5.09Thus, the magnitude of the given vector is 5.09 unit.
Example 5: 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
= 3.6Thus, the magnitude of the given vector is 3.6 unit.
Example 6: 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 = 3.74Thus, the magnitude of the given vector is 3.74 unit.