If |a+b|=|a-b|,then | vedclass

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(D) Given the equation $|a+b|=|a-b|$.Squaring both sides,we get $|a+b|^2 = |a-b|^2$.Using the property $|x|^2 = x \cdot x$,we have $(a+b) \cdot (a+b)

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$a$ and $b$ are perpendicular.

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(D) Given the equation $|a+b|=|a-b|$.Squaring both sides,we get $|a+b|^2 = |a-b|^2$.Using the property $|x|^2 = x \cdot x$,we have $(a+b) \cdot (a+b) = (a-b) \cdot (a-b)$.Expanding the dot products: $a \cdot a + b \cdot b + 2(a \cdot b) = a \cdot a + b \cdot b - 2(a \cdot b)$.Simplifying the equation: $2(a \cdot b) = -2(a \cdot b)$,which implies $4(a \cdot b) = 0$.Therefore,$a \cdot b = 0$.Since the dot product of two non-zero vectors is zero if and only if they are perpendicular,$a$ and $b$ are perpendicular.

If $|a+b|=|a-b|$,then

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