If |a + b| |a - b|,then the angle between a | vedclass

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(A) Given the inequality $|a + b| > |a - b|$.Squaring both sides,we get:$|a + b|^2 > |a - b|^2$$(a + b) \cdot (a + b) > (a - b) \cdot (a - b)$$|a|^2 +

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(A) Given the inequality $|a + b| > |a - b|$.Squaring both sides,we get:$|a + b|^2 > |a - b|^2$$(a + b) \cdot (a + b) > (a - b) \cdot (a - b)$$|a|^2 + |b|^2 + 2(a \cdot b) > |a|^2 + |b|^2 - 2(a \cdot b)$$2(a \cdot b) > -2(a \cdot b)$$4(a \cdot b) > 0$$a \cdot b > 0$Since $a \cdot b = |a||b| \cos 	heta$,we have $|a||b| \cos 	heta > 0$.Since the magnitudes $|a|$ and $|b|$ are positive,$\cos 	heta > 0$.This implies that the angle $	heta$ between $a$ and $b$ must be acute,i.e.,$0 \le 	heta < \frac{\pi}{2}$.

If $|a + b| > |a - b|$,then the angle between $a$ and $b$ is

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