vector proof

**hooke** · February 11th 2010, 04:52 AM

(1)Given two vectors , a and b where a and b are both not 0 , show that if |a+b|=|a-b| , then a and b are perpendicular to each other .

(2) if a and b are perpendicular to each other , show that

|a+b|=|a-b|= $\sqrt{|<b>a</b>|^2+|<b>b</b>|^2}$

**earboth** · February 11th 2010, 08:07 AM

Originally Posted by hooke

(1)Given two vectors , a and b where a and b are both not 0 , show that if |a+b|=|a-b| , then a and b are perpendicular to each other .

(2)...

Square both sides of the equation to get rid of the absolute values:

$|\vec a + \vec b | = |\vec a - \vec b | ~\implies~(\vec a)^2 + 2 \vec a \vec b + (\vec b)^2 = (\vec a)^2 - 2 \vec a \vec b + (\vec b)^2$

Collect like terms:

$2 \vec a \vec b = -2 \vec a \vec b~\implies~4 \vec a \vec b = 0$

Thus $\vec a \vec b = 0$ which means $\vec a \perp \vec b$

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