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

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(A) Given vectors are $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} - 2\hat{k}$.First, calculate the dot product: $\vec{

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

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(A) Given vectors are $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} - 2\hat{k}$.First, calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (-2)(-2) = 2 - 2 + 4 = 4$.Calculate the magnitudes squared: $|\vec{a}|^2 = 1^2 + 2^2 + (-2)^2 = 1 + 4 + 4 = 9$ and $|\vec{b}|^2 = 2^2 + (-1)^2 + (-2)^2 = 4 + 1 + 4 = 9$.The orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} = \frac{4}{9} \vec{b}$.The orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a} = \frac{4}{9} \vec{a}$.Now, $|\vec{x} - \vec{y}| = |\frac{4}{9} \vec{b} - \frac{4}{9} \vec{a}| = \frac{4}{9} |\vec{b} - \vec{a}|$.Calculate $\vec{b} - \vec{a} = (2-1)\hat{i} + (-1-2)\hat{j} + (-2 - (-2))\hat{k} = \hat{i} - 3\hat{j} + 0\hat{k}$.The magnitude $|\vec{b} - \vec{a}| = \sqrt{1^2 + (-3)^2 + 0^2} = \sqrt{1 + 9} = \sqrt{10}$.Therefore, $|\vec{x} - \vec{y}| = \frac{4}{9} \sqrt{10}$.

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

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