### Q.6:- **Establish the following vector inequalities geometrically or otherwise:**

(a) |**a ***+ ***b**| ≤ |**a**| + |**b**|

(b) |**a ***+ ***b**| ≥ ||**a**| − |**b**||

(c) |**a ***− ***b**| ≤ |**a**| + |**b**|

(d) |**a ***− ***b**| ≥ ||**a**| − |**b**||

When does the equality sign above apply?

**Answer:-**

(a) Let two vectors a and b be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we can write:

**OM** = | **a** | …(i)

**MN** = **OP** = | **b** | ….(ii)

**ON **= | **a + b** | …..(iii)

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in ΔOMN, we have:

**ON < **(**OM + MN**)

| **a + b** | < |** a | + | b **| ….(iv)

If the two vectors a and b act along a straight line in the same direction, then we can write:

| **a + b | = | a | + | b |** ….. (v)

Combining equations (*iv*) and (*v*), we get:

**| a + b | ****≤ | a | + | b** |

(b) Let two vectors a and b be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we have:

**| OM | = | a |**

**| MN | = | OP | = | b |**

**| ON | = | a + b |**

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in ΔOMN, we have:

**ON + MN > OM**

**ON + OM > MN**

**| ON | > | OM – OP | (∵ OP = MN)**

**| a + b | > | | a | – | b | | ** ….(iv)

If the two vectors a and b act along a straight line in the same direction, then we can write:

**| a + b | = ****| | a | – | b | | ** …..(v)

Combining equations (*iv*) and (*v*), we get:

**| a + b | ≥ ****| | a | – | b | | **

(c) Let two vectors a and b be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

Here we have:

**| OR | = | PS | = | b |** …(i)

**| OP | = | a | ** ….(ii)

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have:

**OS < OP + PS**

**| a – b | < | a | + | -b |**

**| a – b | < | a | + | b | ** … (iii)

If the two vectors act in a straight line but in opposite directions, then we can write:

**| a – b | = | a | + | b | ** … (iv)

Combining equations (*iii*) and (*iv*), we get:

**| a – b | ≤ | a | + | b | **

(d) Let two vectors a and b be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

**OS + PS > OP** …..(i)

**OS > OP – PS** ….(ii)

**| a – b | > | a | – | b |** ….(iii)

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:

**| | a – b | | > | | a | – | b | |**

**| a – b | > | | a | – | b | |** ….(iv)

If the two vectors act in a straight line but in the opposite directions, then we can write:

**| a – b | = | | a | – | b | | ** ….(v)

Combining equations (*iv*) and (*v*), we get:

**| a – b | ≥ | | a | – | b | | **