Establish the following inequalities geometrically or otherwise ,
(a) ` | vec A + vec B| lt= | vec A | + |vec B|` ,
(b) ` | vec A + vec B| gt= ||vec A|- | vec B||`
(C )` | vec A- vec B| lt= | vec A| + |vec B|`
(d) `| vec A- vec B| gt= || vec A| - | vec B||`
When does the equality sign above apply ?

Explanation

Consider two vectors `vecA` and `vecB` be represented by the sides `vec(OP)` and `vec(OQ)` of a parallelogram OPSQ. According to parallelogram law of vector addition, `(vecA+vecB)` will be represented by `vec(OS)` as shown in Fig. Thus,
OP = `|vecA|`, `OQ=PS=|vecB|`
and OS = `|verA+verB|`
a) To prove `|vecA+vecB| larr|vecA|+|vecB|`
We know that the length of one side of a triangle is always less than the sum of the lengths of the other two sides. Hence from `Delta` OPS, we have,
OSltOP+PS or OSltOP+OQ or `|vecA+vecB|lt |vecA|+|vecB|`..............(i)
If the two vectors `vecA` and `vecB` are acting along the same straight line and in the same direction
then, `|vecA + vecB|=|vecA|+|vecB|`..............(ii)
Combining the conditions mentioned in (i) and (ii) we have
`|vecA+vecB| larr|vecA|+|vecB|`
b) To prove `|vecA+vecB|gt-||vecA|-|vecB||`
From `Delta`OPS, we have `OS+PSgtOP` or `OSgt|OP-PS|` or `OSgt|OP-OQ|`............(iii) `(therefore` PS=OQ)
The modules of (OP-PS) has been taken because the L.H.S is always positive but the R.H.S may be negative if `OPltPS`. thus from (iii) , we have,
`|vecA+vecB| gt ||vecA|-|vecB||` .............(iv)
if the two vectors `vecA` and `vecB` are acting along a straight line in opposite directions, then
`|vecA+vecB|=||vecA|-|vecB||`............(v)
Combining the conditions mentioned in (iv) and (v), we get
`|vecA+vecB|ge||vecA|-|vecB||`
c) To prove `|vecA-vecB|le|vecA|+|vecB|`
In fig. `vecOL` and `vec(OM)` represents vectors `vecA` and `vecB` respectively. Here `vec(ON)` represents `vec(A-B)`.
consider the `Delta` OMN,
`ONltMN+OM`
`|vecA-vecB|lt|vecA|+|-vecB|`............(vi)
When `vecA` and `vecB` are along the same line, but points in the opposite direction, then
When `vecA` and `vecB` are along the same straight line, but points in the opposite direction, then
`|vecA-vecB|= |vecA|+|vecB|`............(vii)
combining equation (vi) and (vii), we get
`|vecA-vecB| larr|vecA|+|vecB|`
d) To prove `|vecA-vecB| le |vecA|+|vecB|`
Let us consider the `Delta OMN`,
`ON+OMgtMN` or `ONgt[MN-OM]`
since, `MN=OL` `therefore` `ONgt|OL-OM|`
or `|vecA-vecB| gt ||vecA|-|vecB||`...................(viii)
When `vecA` and `vecB` are along the same straight line and point in the same direction,then
`|vecA-vecB|= |vecA|-|vecB|`............(ix)
Combining equations (viii) and (ix), we get
`|vecA-vecB|ge||vecA|-|vecB||`