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 +5jk 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 ba:

(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.

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