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∣


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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 byTANMAY PANACEA last updated on 16/Oct/20

	$m o s t w e l c o m e s i r$
	Commented byTANMAY PANACEA last updated on 16/Oct/20

	$y e s s i r . . .$
	Commented by1549442205PVT last updated on 16/Oct/20

	$Thank sir . You are welcome .$ $You corrected exactly . When$ $θ = π then there two possibilities$ $∣ a + b ∣=∣ a - b ∣ = 0 o r ∣ a + b ∣<∣ a - b ∣$ $Question means the inequalities are$ $really true .$
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