The condition ( a cdot b )^2 = a^2 b^2 is sat | vedclass

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(A) The dot product of two vectors $a$ and $b$ is defined as $a \cdot b = |a| |b| \cos 	heta$,where $	heta$ is the angle between the vectors.Squarin

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$a$ is parallel to $b$

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(A) The dot product of two vectors $a$ and $b$ is defined as $a \cdot b = |a| |b| \cos 	heta$,where $	heta$ is the angle between the vectors.Squaring both sides,we get $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 	heta$.Given the condition $(a \cdot b)^2 = a^2 b^2$,we substitute the expression:$|a|^2 |b|^2 \cos^2 	heta = |a|^2 |b|^2$.Assuming $a$ and $b$ are non-zero vectors,we divide by $|a|^2 |b|^2$ to get $\cos^2 	heta = 1$.This implies $\cos 	heta = \pm 1$,which means $	heta = 0^{\circ}$ or $	heta = 180^{\circ}$.In both cases,the vectors $a$ and $b$ are parallel (or anti-parallel). Thus,the condition is satisfied when $a$ is parallel to $b$.

The condition $( a \cdot b )^2 = a^2 b^2$ is satisfied when

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