What-condition-should-be-satisfied-by-the-vectors-a-and-b-for-the-following-relations-to-hold-true-a-a-b-a-b-b-a-b-gt-a-b-c-a-b-lt-a-b-

Question Number 118286 by 1549442205PVT last updated on 16/Oct/20

What condition should be satisfied by

the vectors a and b for the following

relations to hold true : (a) ∣ a + b ∣=∣ a - b ∣

; (b) ∣ a + b ∣>∣ a - b ∣; (c) ∣ a + b ∣< ∣ a - b ∣

Answered by TANMAY PANACEA last updated on 16/Oct/20

a) ∣ a + b ∣=∣ a - b ∣

∣ a + b ∣^{2} =∣ a - b ∣^{2}

(a + b) . (a + b) = (a - b) . (a - b)

a^{2} + b^{2} + 2 a b c o s θ = a^{2} + b^{2} - 2 a b c o s θ

[a . b = a b c o s θ a . a = a^{2} b . b = b^{2}]

4 a b c o s θ = 0 θ = \frac{π}{2} s o a ⊥ b

Answered by TANMAY PANACEA last updated on 16/Oct/20

b) s i m i l a r l y

∣ a + b ∣>∣ a - b ∣

\sqrt{a^{2} + b^{2} + 2 a b c o s θ} > \sqrt{a^{2} + b^{2} - 2 a b c o s θ}

c o s θ > 0 s o θ = a c u t e a n g l e

\frac{π}{2} > θ > 0

c) a + b ∣<∣ a - b ∣

\sqrt{a^{2} + b^{2} + 2 a b c o s θ} < \sqrt{a^{2} + b^{2} - 2 a b c o s θ}

c o s θ = - v e

c o s θ < 0

π > θ > \frac{π}{2} c o r r e c t i o n d o n e

p l z c h e c k

Commented by TANMAY PANACEA last updated on 16/Oct/20

m o s t w e l c o m e s i r

Commented by TANMAY PANACEA last updated on 16/Oct/20

y e s s i r \dots

Commented by 1549442205PVT last updated on 16/Oct/20

Thank sir . You are welcome .

You corrected exactly . When

θ = π then there two possibilities

∣ a + b ∣=∣ a - b ∣ = 0 o r ∣ a + b ∣<∣ a - b ∣

Question means the inequalities are

really true .