Let a=hat{i}+2 hat{j}-2 hat{k} and b=2 hat{i} | vedclass

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(C) The orthogonal projection of $a$ on $b$ is given by $x = \frac{a \cdot b}{|b|^2} b$. The orthogonal projection of $b$ on $a$ is given by $y = \fra

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$\frac{4}{9} \sqrt{10}$

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(C) The orthogonal projection of $a$ on $b$ is given by $x = \frac{a \cdot b}{|b|^2} b$. The orthogonal projection of $b$ on $a$ is given by $y = \frac{a \cdot b}{|a|^2} a$. Given $a = \hat{i} + 2\hat{j} - 2\hat{k}$ and $b = 2\hat{i} - \hat{j} - 2\hat{k}$. Calculate the dot product: $a \cdot b = (1)(2) + (2)(-1) + (-2)(-2) = 2 - 2 + 4 = 4$. Calculate the magnitudes squared: $|a|^2 = 1^2 + 2^2 + (-2)^2 = 1 + 4 + 4 = 9$ and $|b|^2 = 2^2 + (-1)^2 + (-2)^2 = 4 + 1 + 4 = 9$. Thus,$x - y = \frac{a \cdot b}{|b|^2} b - \frac{a \cdot b}{|a|^2} a = \frac{4}{9} b - \frac{4}{9} a = \frac{4}{9} (b - a)$. Calculate $b - a = (2\hat{i} - \hat{j} - 2\hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) = \hat{i} - 3\hat{j}$. Then $x - y = \frac{4}{9} (\hat{i} - 3\hat{j})$. Finally,$|x - y| = \frac{4}{9} \sqrt{1^2 + (-3)^2} = \frac{4}{9} \sqrt{1 + 9} = \frac{4}{9} \sqrt{10}$.

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$,then $|x-y|=$

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