If vectors satisfy the condition |a - c| = |b | vedclass

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

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(A) Given the condition $|a - c| = |b - c|$.Squaring both sides,we get $|a - c|^2 = |b - c|^2$.Using the property $|x|^2 = x \cdot x$,we have $(a - c) \cdot (a - c) = (b - c) \cdot (b - c)$.$a \cdot a - 2(a \cdot c) + c \cdot c = b \cdot b - 2(b \cdot c) + c \cdot c$.$|a|^2 - 2(a \cdot c) = |b|^2 - 2(b \cdot c)$.Now,consider the expression $E = (b - a) \cdot \left( c - \frac{a + b}{2} \right)$.$E = (b - a) \cdot \left( \frac{2c - a - b}{2} \right) = \frac{1}{2} (b - a) \cdot (2c - a - b)$.$E = \frac{1}{2} [2(b \cdot c) - (b \cdot a) - |b|^2 - 2(a \cdot c) + |a|^2 + (a \cdot b)]$.Since $a \cdot b = b \cdot a$,the terms cancel out.$E = \frac{1}{2} [2(b \cdot c) - 2(a \cdot c) - |b|^2 + |a|^2]$.From our initial condition,$|a|^2 - |b|^2 = 2(a \cdot c) - 2(b \cdot c)$.Substituting this into the expression for $E$:$E = \frac{1}{2} [2(b \cdot c) - 2(a \cdot c) + (2(a \cdot c) - 2(b \cdot c))] = 0$.

If vectors satisfy the condition $|a - c| = |b - c|$,then $(b - a) \cdot \left( c - \frac{a + b}{2} \right)$ is equal to

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