NCERT Solutions
Class 11 Physics
Motion in a Plane
Q. 3.6
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 do the equality sign above apply?
- Let two vectors a ⃗ and b ⃗ be represented by the adjacent sides of a parallelogram OMNP, as shown in the figure.
| 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 a⃗ and b⃗ 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 ⃗|
- Let two vectors a⃗ and b⃗ be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.
| 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 +MN) > OM
(ON + OM) > MN
| ON ⃗| > | OM ⃗ - OP ⃗| (because OP=MN)
| a ⃗ + b ⃗ | > || a ⃗| - | b ⃗|| (iv)
If the two vectors a⃗ and b⃗ 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 a⃗ and b⃗ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
| 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,
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 a⃗ and b⃗ be represented by the adjacent sides of a parallelogram PORS, as shown in the 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 ⃗||