for a 2d vectors if ∣a + b∣ = ∣a−b∣ what relationship does a and b have?


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Question Number 92465 by Rio Michael last updated on 07/May/20

$for a 2 d vectors if ∣ a + b ∣ = ∣ a - b ∣ what relationship does a and b have ?$
Commented by mr W last updated on 07/May/20

$a ⊥ b$
Commented by Rio Michael last updated on 07/May/20

$thanks sir, any prove for that sir ?$
Commented by mr W last updated on 07/May/20

$∣ a + b ∣ a n d ∣ a - b ∣ a r e t h e b o t h d i a g o n a l s$ $o f t h e p a r a l l e o g r a m w i t h a a n d b a s$ $s i d e s . w h e n b o t h d i a g o n a l s s h o u l d b e$ $e q u a l, t h e p a r a l l e l o g r a m m u s t b e a$ $r e c t a n g l e, i . e . a a n d b m u s t b e ⊥ t o$ $e a c h o t h e r .$
Commented by mr W last updated on 13/May/20

Commented by Rio Michael last updated on 07/May/20

$following your explanation i checked that$ $thank you sir .$
Commented by Prithwish Sen 1 last updated on 07/May/20

$∵ a + b and a - b are the two diagonals of the$ $parallalogram formed by the vectors a and b$ $as the two adjacent sides then if ∣ a + b ∣=∣ a - b ∣$ $only implies that the diagonals are same and it$ $can only happen when the parallalogram so$ $formed by a and b is a rectangle . And that is why$ $the two adjacent sides must be ⊥ and i . e$ $a ⊥ b .$
Commented by Rio Michael last updated on 07/May/20

$another good explanation thanks$
	Answered by behi83417@gmail.com last updated on 07/May/20
	$∣a+b∣=(√(a^2 +b^2 −2abcosθ)) ∣a−b∣=(√(a^2 +b^2 +2abcosθ)) [θ=angle between a and b] ∣a+b∣=∣a−b∣⇒ a^2 +b^2 −2abcosθ=a^2 +b^2 +2abcosθ ⇒4abcosθ=0⇒ { ((a=0 (or))),((b=0 (or))),((cosθ=0⇒θ=(π/2)⇒a⊥b)) :}$
	$∣ a + b ∣= \sqrt{a^{2} + b^{2} - 2 abcos θ}$ $∣ a - b ∣= \sqrt{a^{2} + b^{2} + 2 abcos θ}$ $[θ = angle between a and b]$ $∣ a + b ∣=∣ a - b ∣\Rightarrow$ $a^{2} + b^{2} - 2 abcos θ = a^{2} + b^{2} + 2 abcos θ$ $\Rightarrow 4 abcos θ = 0 \Rightarrow {\begin{cases} a = 0 (or) \\ b = 0 (or) \\ \cos θ = 0 \Rightarrow θ = \frac{π}{2} \Rightarrow a ⊥ b \end{cases}$
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