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(B) Given that $\frac{1}{2}|a-b|=\sin(\lambda 	heta)$.Squaring both sides,we get $\frac{1}{4}|a-b|^2 = \sin^2(\lambda 	heta)$.Since $a$ and $b$ are

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(B) Given that $\frac{1}{2}|a-b|=\sin(\lambda 	heta)$.Squaring both sides,we get $\frac{1}{4}|a-b|^2 = \sin^2(\lambda 	heta)$.Since $a$ and $b$ are unit vectors,$|a|=1$ and $|b|=1$.$|a-b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) = 1 + 1 - 2|a||b|\cos 	heta = 2 - 2\cos 	heta$.Substituting this into the equation: $\frac{1}{4}(2 - 2\cos 	heta) = \sin^2(\lambda 	heta)$.$\frac{1}{2}(1 - \cos 	heta) = \sin^2(\lambda 	heta)$.Using the identity $1 - \cos 	heta = 2\sin^2(\frac{	heta}{2})$,we get $\frac{1}{2}(2\sin^2(\frac{	heta}{2})) = \sin^2(\lambda 	heta)$.$\sin^2(\frac{	heta}{2}) = \sin^2(\lambda 	heta)$.Comparing the arguments,$\lambda 	heta = \frac{	heta}{2}$,which implies $\lambda = \frac{1}{2}$.Therefore,$4\lambda^2 = 4(\frac{1}{2})^2 = 4(\frac{1}{4}) = 1$.

Let $a$ and $b$ be unit vectors with $\theta$ as the acute angle between them. If $\frac{1}{2}|a-b|=\sin(\lambda \theta)$,then $4 \lambda^2=$

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