If a neq 0, b neq 0 and |a + b| = |a - b|,the | vedclass

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(B) Given $|a + b| = |a - b|$.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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Perpendicular to each other

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(B) Given $|a + b| = |a - b|$.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 product: $a \cdot a + 2(a \cdot b) + b \cdot b = a \cdot a - 2(a \cdot b) + b \cdot b$.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,the vectors $a$ and $b$ are perpendicular to each other.

If $a \neq 0, b \neq 0$ and $|a + b| = |a - b|$,then the vectors $a$ and $b$ are . . . .

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