Vectors Problem Example 2

Description:

Problem

Establish the following vector inequalities geometrically or otherwise −

1. |a→ + b→| ≤ |a→| + |b→|

2. |a→ + b→| ≥ |a→| - |b→|

Solution

For ease of notation, |a→| will be represented by ‘a’ and |b→| will be represented by ‘b’.

Part 1 – Analytical Method

• a→ and b→ are two arbitary vectors.

• The angle between them is θ, it can be acute or obtuse.

Now, we know,

|a→ + b→| = √a² + b² + 2ab cosθ

Max. value of cosθ = 1. (when θ = 0^o or 0 rad)

• Hence, |a→ + b→| ≤ √a² + b² + 2ab

• Or, |a→ + b→| ≤ √(a + b)²

• Or, |a→ + b→| ≤ (|a→| + |b→|) ..... Proved

|a→ + b→| = (|a→| + |b→|) only when cosθ = 1, else it is always, |a→ + b→| < (|a→| + |b→|)

NOTE − a and |a→| are used interchangeably. Same for b and |b→|.

Part 1 – Geometrical Method

• a→ and b→ are two arbitary vectors.

• The angle between them is θ, it can be acute or obtuse.

• Taking S→ = a→ + b→, case by case each time changing the value of angle θ.

• Clearly, |a→ + b→| is maximum when the angle θ = 0 rad. Hence,

|a→ + b→| ≤ |a→| + |b→|

Part 2 – Analytical Method

• a→ and b→ are two arbitary vectors.

• The angle between them is θ, it can be acute or obtuse.

Now, we know,

|a→ + b→| = √a² + b² + 2ab cosθ

Max. value of cosθ = 1. (when θ = 0^o or 0 rad)

• Hence, |a→ + b→| ≤ √a² + b² + 2ab

• Or, |a→ + b→| ≤ √(a + b)²

• Or, |a→ + b→| ≤ (|a→| + |b→|) ..... Proved

|a→ + b→| = (|a→| + |b→|) only when cosθ = 1, else it is always, |a→ + b→| < (|a→| + |b→|)

NOTE − a and |a→| are used interchangeably. Same for b and |b→|.

Part 2 – Geometrical Method

• a→ and b→ are two arbitary vectors.

• The angle between them is θ, it can be acute or obtuse.

• Taking S→ = a→ + b→, case by case each time changing the value of angle θ.

• Clearly, |a→ + b→| is maximum when the angle θ = π/2 rad. Hence,

|a→ + b→| ≤ |a→| + |b→|