Let x = hat{i} + hat{j} and y = 3hat{i} - 2ha | vedclass

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(C) Given, $x = \hat{i} + \hat{j}$ and $y = 3\hat{i} - 2\hat{k}$.The conditions are $r 	imes x = y 	imes x$ and $r 	imes y = x 	imes y$.From $r

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$4\hat{i} + \hat{j} - 2\hat{k}$

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(C) Given, $x = \hat{i} + \hat{j}$ and $y = 3\hat{i} - 2\hat{k}$.The conditions are $r 	imes x = y 	imes x$ and $r 	imes y = x 	imes y$.From $r 	imes x = y 	imes x$, we have $(r - y) 	imes x = 0$, which implies that $(r - y)$ is parallel to $x$.Thus, $r - y = \lambda x$, or $r = y + \lambda x$.Substituting the vectors: $r = (3\hat{i} - 2\hat{k}) + \lambda(\hat{i} + \hat{j}) = (3 + \lambda)\hat{i} + \lambda\hat{j} - 2\hat{k}$.Given the magnitude $|r| = \sqrt{21}$, we have $|r|^2 = 21$.$(3 + \lambda)^2 + \lambda^2 + (-2)^2 = 21$.$9 + 6\lambda + \lambda^2 + \lambda^2 + 4 = 21$.$2\lambda^2 + 6\lambda + 13 = 21 \Rightarrow 2\lambda^2 + 6\lambda - 8 = 0$.Dividing by $2$: $\lambda^2 + 3\lambda - 4 = 0$.$(\lambda + 4)(\lambda - 1) = 0$, so $\lambda = 1$ or $\lambda = -4$.For $\lambda = 1$, $r = (3 + 1)\hat{i} + 1\hat{j} - 2\hat{k} = 4\hat{i} + \hat{j} - 2\hat{k}$.Checking the second condition $r 	imes y = x 	imes y$: $(r - x) 	imes y = 0$, so $r - x$ must be parallel to $y$.For $r = 4\hat{i} + \hat{j} - 2\hat{k}$, $r - x = (4\hat{i} + \hat{j} - 2\hat{k}) - (\hat{i} + \hat{j}) = 3\hat{i} - 2\hat{k} = y$. Since $y$ is parallel to $y$, this is correct.

Let $x = \hat{i} + \hat{j}$ and $y = 3\hat{i} - 2\hat{k}$. Then, the vector $r$ of magnitude $\sqrt{21}$ satisfying $r \times x = y \times x$ and $r \times y = x \times y$ is

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