Question

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 1

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

Here, we can write: 
|OM| =|a| ... (1) 
|MN|=|OP|=6 ...(ii)
ON |= |a+b| ... (iii)

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in OMN, we have: ON < x`(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|

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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 | ... (i)
| MN |= |OP| =|b| ... (ii)
|ON |= |a + b| ... (ii)

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

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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 OPS,
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 direction, then we can write :
|a-b| = |a| + |b| ....(iv)

Combining equations (iii) and (iv), we get:
|a-b| ≤|a| + |b|

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

The following relations can be written for the given parallelogram. 
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||