Question

Mod of a vector + b vector less than equal to mod of a vector + mod of b vector

Answer 1

Answer:

[mod(a-b)]^2 =(a-b).(a-b) = mod(a)^2 + mod(b)^2 -2(a.b)

= mod(a)^2 + mod(b)^2 - 2( mod a)(mod b)cos t (where t is the angle between

a & b) is less or equal to [(mod a)^2 +(mod b)^2 + 2 (mod a)(mod b)] (since

-cos t is less or equal to 1) = [mod a + mod b]^2. Upon taking +ve square root

we get that mod(a-b) less or equal mod a +mod b. Equality happens if a and

b are directed opposite to each other, for then t = (pi) and so cos t = -1.

Explanation:

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