Let a = sin^2 x hat{i} + cos^2 x hat{j} + hat | vedclass

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(B) Given $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$.Area of a parallelogram with adjacent sides $u$ and $v$ is $|u 	imes v|$.$1$. Area of 1

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$[3, 4]$

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(B) Given $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$.Area of a parallelogram with adjacent sides $u$ and $v$ is $|u 	imes v|$.$1$. Area of 1st parallelogram: $|a 	imes \hat{i}| = |(\sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}) 	imes \hat{i}| = |-\cos^2 x \hat{k} + \hat{j}| = \sqrt{\cos^4 x + 1}$.So, $|a 	imes \hat{i}|^2 = \cos^4 x + 1$.$2$. Area of 2nd parallelogram: $|a 	imes \hat{j}| = |(\sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}) 	imes \hat{j}| = |\sin^2 x \hat{k} - \hat{i}| = \sqrt{\sin^4 x + 1}$.So, $|a 	imes \hat{j}|^2 = \sin^4 x + 1$.$3$. Area of 3rd parallelogram: $|a 	imes \hat{k}| = |(\sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}) 	imes \hat{k}| = |-\sin^2 x \hat{j} + \cos^2 x \hat{i}| = \sqrt{\sin^4 x + \cos^4 x}$.So, $|a 	imes \hat{k}|^2 = \sin^4 x + \cos^4 x$.Sum of squares of areas $A = |a 	imes \hat{i}|^2 + |a 	imes \hat{j}|^2 + |a 	imes \hat{k}|^2 = (\cos^4 x + 1) + (\sin^4 x + 1) + (\sin^4 x + \cos^4 x) = 2 + 2(\sin^4 x + \cos^4 x)$.Using $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x)$, we get:$A = 2 + 2(1 - \frac{1}{2}\sin^2(2x)) = 4 - \sin^2(2x)$.Since $0 \leq \sin^2(2x) \leq 1$, we have $3 \leq 4 - \sin^2(2x) \leq 4$.Thus, $A \in [3, 4]$.

Let $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$, where $x \in R$. If the pairs of vectors $(a, \hat{i})$, $(a, \hat{j})$, and $(a, \hat{k})$ are adjacent sides of $3$ distinct parallelograms and $A$ is the sum of the squares of the areas of these parallelograms, then $A$ lies in the interval

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