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 | |