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

Vedclass · Accepted Answer

$a$ is parallel to $b$

Vedclass · Answer

(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 $b$.

Column $I$	Column $II$
$(A)$ Component along $x$-axis	$(p)$ $5 \text{ unit}$
$(B)$ Component along vector $(2 \hat{i} + \hat{j} + 2 \hat{k})$	$(q)$ $4 \text{ unit}$
$(C)$ Component along $(6 \hat{i} + 8 \hat{j} - 10 \hat{k})$	$(r)$ $0$
$(D)$ Component along $(-3 \hat{i} - 4 \hat{j} + 5 \hat{k})$	$(s)$ None

Column $I$	Column $II$
$(A)$ $\theta=60^{\circ}$	$(p)$ $n=\sqrt{3}$
$(B)$ $\theta=90^{\circ}$	$(q)$ $n=1$
$(C)$ $\theta=120^{\circ}$	$(r)$ $n=\sqrt{2}$
$(D)$ $\theta=180^{\circ}$	$(s)$ $n=2$

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

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