How to work out the magnitude (modulus) of a vector in the form xi + yj + zk.
To work out the magnitude (or modulus) of a vector in the form xi + yj +zk all you need to do is square the coefficients of i, j and k, add these squares together and square root the answer:
√(x²+y²+z²)
Example 1
If a = 3i + 5j – 3k, work out |a|.
|a| stands for the magnitude of the vector, so substitute x = 3, y = 5 and z = -3 into the above formula:
|a| = √(x²+y²+z²)
= √(3² + 5² + (-3)²)
= √(9 + 25 + 9)
= √43
Example 2
If b = 2i + 9j – 4k, work out |b|.
|b| stands for the magnitude of the vector, so substitute x = 2, y = 9 and z = -4 into the above formula:
|b| = √(x²+y²+z²)
= √(2² + 9² + (-4)²)
= √(4 + 81 + 16)
= √101
Example 3
If c = 10i + 4j – 7k, work out |c|.
|c| stands for the magnitude of the vector, so substitute x = 10, y = 4 and z = -7 into the above formula:
|c| = √(x²+y²+z²)
= √(10² + 4² + (-7)²)
= √(100 + 16 + 49)
= √165
Example 4
If a = 3i – 5j – 7k and b = 2i +5j –k work out |AB|. Round your answer off to 1 decimal place.
Before you can work out the magnitude you will need to work out the vector AB:
To find AB work out b –a:
(3i -5j -7k) – (2i +5j-k) = i – 10j -6k
So AB = i – 10j -6k
All you need to do next is work out the magnitude of AB. The coefficient of x,y and z are 1, 10 and -6:
|AB| = √(1² + 10² + (-6)²) = √137
= 11.7 to 1 decimal place.