For any vector x, where hat{i}, hat{j}, hat{k | vedclass

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(B) Let $x = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.Then, $x 	imes \hat{i} = (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) 	imes \hat{

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$2|x|^{2}$

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(B) Let $x = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.Then, $x 	imes \hat{i} = (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) 	imes \hat{i} = -\beta \hat{k} + \gamma \hat{j}$.Similarly, $x 	imes \hat{j} = \alpha \hat{k} - \gamma \hat{i}$ and $x 	imes \hat{k} = -\alpha \hat{j} + \beta \hat{i}$.Now, $(x 	imes \hat{i})^{2} = (x 	imes \hat{i}) \cdot (x 	imes \hat{i}) = (-\beta \hat{k} + \gamma \hat{j}) \cdot (-\beta \hat{k} + \gamma \hat{j}) = \beta^{2} + \gamma^{2}$.Similarly, $(x 	imes \hat{j})^{2} = \alpha^{2} + \gamma^{2}$ and $(x 	imes \hat{k})^{2} = \alpha^{2} + \beta^{2}$.Adding these, we get $(x 	imes \hat{i})^{2} + (x 	imes \hat{j})^{2} + (x 	imes \hat{k})^{2} = (\beta^{2} + \gamma^{2}) + (\alpha^{2} + \gamma^{2}) + (\alpha^{2} + \beta^{2}) = 2(\alpha^{2} + \beta^{2} + \gamma^{2})$.Since $|x|^{2} = \alpha^{2} + \beta^{2} + \gamma^{2}$, the expression equals $2|x|^{2}$.

For any vector $x$, where $\hat{i}, \hat{j}, \hat{k}$ have their usual meanings, the value of $(x \times \hat{i})^{2} + (x \times \hat{j})^{2} + (x \times \hat{k})^{2}$ is equal to

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