Key Concepts and Formulas

1. Magnitude of a Vector Sum: For any two vectors $\vec{a}$ and $\vec{b}$, the square of the magnitude of their sum is given by: $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$ This identity arises from the property of the dot product: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$ since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.

2. Lagrange's Identity: For any two vectors $\vec{a}$ and $\vec{b}$, the following relationship holds: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ This identity connects the magnitude of the cross product and the dot product of two vectors with their individual magnitudes. It is derived from the definitions $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$.

Step-by-Step Solution

Step 1: Use the first given equation to find the value of $|\vec{b}|^2$. We are given the equation $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$. Using the formula for the magnitude of a vector sum, we expand the left side: $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^{2}+2|\vec{b}|^{2}$ We are also given that $\vec{a} \cdot \vec{b} = 3$. Substituting this into the equation: $|\vec{a}|^2 + |\vec{b}|^2 + 2(3) = |\vec{a}|^{2}+2|\vec{b}|^{2}$ $|\vec{a}|^2 + |\vec{b}|^2 + 6 = |\vec{a}|^{2}+2|\vec{b}|^{2}$ Now, we simplify by subtracting $|\vec{a}|^2$ from both sides: $|\vec{b}|^2 + 6 = 2|\vec{b}|^2$ To solve for $|\vec{b}|^2$, we rearrange the terms: $6 = 2|\vec{b}|^2 - |\vec{b}|^2$ $6 = |\vec{b}|^2$ Thus, we find that $|\vec{b}|^2 = 6$.

Step 2: Use Lagrange's Identity and the remaining information to find $|\vec{a}|^2$. We have the following information:

• $|\vec{b}|^2 = 6$ (from Step 1)
• $\vec{a} \cdot \vec{b} = 3$ (given)
• $|\vec{a} \times \vec{b}|^2 = 75$ (given)

We apply Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ Substitute the known values into the identity: $75 = |\vec{a}|^2 (6) - (3)^2$ $75 = 6|\vec{a}|^2 - 9$ Now, we solve for $|\vec{a}|^2$. First, add 9 to both sides of the equation: $75 + 9 = 6|\vec{a}|^2$ $84 = 6|\vec{a}|^2$ Finally, divide by 6 to find $|\vec{a}|^2$: $|\vec{a}|^2 = \frac{84}{6}$ $|\vec{a}|^2 = 14$

Common Mistakes & Tips

• Identity Recall: Ensure you have memorized the identities for the magnitude of a vector sum and Lagrange's Identity. They are fundamental tools.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with squares and subtractions. A small error can lead to a significantly wrong answer.
• Distinguishing Scalars and Vectors: Remember that $|\vec{a}|^2$ and $(\vec{a} \cdot \vec{b})^2$ are scalar quantities, while $\vec{a}$ and $\vec{b}$ are vectors.

Summary

The problem required the application of two key vector identities. First, the expansion of $|\vec{a}+\vec{b}|^2$ was used in conjunction with the given information to determine $|\vec{b}|^2$. Subsequently, Lagrange's Identity was employed, substituting the known values of $|\vec{b}|^2$, $\vec{a} \cdot \vec{b}$, and $|\vec{a} \times \vec{b}|^2$ to solve for the desired quantity, $|\vec{a}|^2$. The systematic application of these formulas led to the correct result.

The final answer is $\boxed{14}$.