Key Concepts and Formulas

This problem requires a solid understanding of vector algebra, specifically:

1. Projection of a Vector: The projection of vector $\vec{x}$ onto vector $\vec{y}$ is given by: $\text{Proj}_{\vec{y}} \vec{x} = \frac{\vec{x} \cdot \vec{y}}{|\vec{y}|}$
2. Lagrange's Identity: For any two vectors $\vec{a}$ and $\vec{b}$: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$
3. Magnitude of a Difference Vector: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$
4. Properties of Dot Product: $\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x}$ and $\vec{x} \cdot \vec{x} = |\vec{x}|^2$.

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Step-by-Step Solution

We are asked to find the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b}$. This requires us to calculate $\frac{\vec{b} \cdot (\vec{a} - \vec{b})}{|\vec{a} - \vec{b}|}$.

Step 1: Calculate the magnitude of $\vec{a}$ and the magnitude of $\vec{a} \times \vec{b}$. We are given $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$. The magnitude of $\vec{a}$ is: $|\vec{a}| = \sqrt{(1)^2 + (-1)^2 + (2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$ We are given $\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}$. The magnitude of $\vec{a} \times \vec{b}$ is: $|\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (0)^2 + (-1)^2} = \sqrt{4 + 0 + 1} = \sqrt{5}$

Step 2: Determine the magnitude of $\vec{b}$ using Lagrange's Identity. We are given $|\vec{a} \times \vec{b}| = \sqrt{5}$, $|\vec{a}| = \sqrt{6}$, and $\vec{a} \cdot \vec{b} = 3$. Lagrange's Identity states: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. Substituting the known values: $(\sqrt{5})^2 + (3)^2 = (\sqrt{6})^2 |\vec{b}|^2$ $5 + 9 = 6 |\vec{b}|^2$ $14 = 6 |\vec{b}|^2$ $|\vec{b}|^2 = \frac{14}{6} = \frac{7}{3}$

Step 3: Calculate the magnitude of the vector $\vec{a} - \vec{b}$. The magnitude squared of $\vec{a} - \vec{b}$ is given by $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$. Substituting the values we have found: $|\vec{a} - \vec{b}|^2 = (\sqrt{6})^2 + \frac{7}{3} - 2(3)$ $|\vec{a} - \vec{b}|^2 = 6 + \frac{7}{3} - 6$ $|\vec{a} - \vec{b}|^2 = \frac{7}{3}$ Therefore, the magnitude is: $|\vec{a} - \vec{b}| = \sqrt{\frac{7}{3}}$

Step 4: Calculate the dot product $\vec{b} \cdot (\vec{a} - \vec{b})$. Using the distributive property of the dot product: $\vec{b} \cdot (\vec{a} - \vec{b}) = \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b}$ Using the properties $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2$: $\vec{b} \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{b} - |\vec{b}|^2$ Substituting the known values: $\vec{b} \cdot (\vec{a} - \vec{b}) = 3 - \frac{7}{3}$ $\vec{b} \cdot (\vec{a} - \vec{b}) = \frac{9}{3} - \frac{7}{3} = \frac{2}{3}$

Step 5: Calculate the projection of $\vec{b}$ on $\vec{a} - \vec{b}$. The projection is given by $\frac{\vec{b} \cdot (\vec{a} - \vec{b})}{|\vec{a} - \vec{b}|}$. Substituting the values calculated in Steps 3 and 4: $\text{Proj}_{(\vec{a} - \vec{b})} \vec{b} = \frac{\frac{2}{3}}{\sqrt{\frac{7}{3}}}$ To simplify: $= \frac{2}{3} \cdot \frac{1}{\sqrt{\frac{7}{3}}} = \frac{2}{3} \cdot \sqrt{\frac{3}{7}}$ $= \frac{2}{3} \cdot \frac{\sqrt{3}}{\sqrt{7}} = \frac{2 \sqrt{3}}{3 \sqrt{7}}$ Rationalizing the denominator: $= \frac{2 \sqrt{3} \cdot \sqrt{7}}{3 \sqrt{7} \cdot \sqrt{7}} = \frac{2 \sqrt{21}}{3 \cdot 7} = \frac{2 \sqrt{21}}{21}$ Alternatively, we can simplify $\frac{2}{3} \cdot \frac{\sqrt{3}}{\sqrt{7}}$ as: $= \frac{2}{\sqrt{3} \cdot \sqrt{7}} = \frac{2}{\sqrt{21}}$

This matches option (A).

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Common Mistakes & Tips

• Lagrange's Identity is Key: When given both dot and cross product information, this identity is often the most direct path to finding unknown magnitudes.
• Avoid Premature Square Roots: Keep magnitudes squared as fractions (e.g., $|\vec{b}|^2 = \frac{7}{3}$) until they are needed in the final division, as this often simplifies calculations.
• Algebraic Manipulation of Radicals: Be careful when simplifying expressions involving square roots, especially when they are in the denominator.

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Summary

To find the projection of $\vec{b}$ on $\vec{a} - \vec{b}$, we first calculated the magnitudes of $\vec{a}$ and $\vec{a} \times \vec{b}$. Then, using Lagrange's Identity, we determined the magnitude of $\vec{b}$. With $|\vec{a}|$, $|\vec{b}|$, and $\vec{a} \cdot \vec{b}$ known, we found the magnitude of $\vec{a} - \vec{b}$ and the dot product $\vec{b} \cdot (\vec{a} - \vec{b})$. Finally, these values were used in the projection formula to arrive at the answer.

The final answer is $\boxed{\frac{2}{\sqrt{21}}}$.