Key Concepts and Formulas

1. Magnitude of Vector Sum/Difference: For any two vectors $\vec{u}$ and $\vec{v}$,

   • $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$
   • $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u} \cdot \vec{v}$ These formulas are derived from the dot product property $(\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v})$.

2. Componendo and Dividendo Rule: If $\frac{P}{Q} = \frac{R}{S}$, then $\frac{P+Q}{P-Q} = \frac{R+S}{R-S}$. This algebraic rule is useful for simplifying ratios involving sums and differences.

3. Given Condition: $|\vec{a}| = |\vec{b}|$. This implies $|\vec{a}|^2 = |\vec{b}|^2$, which can be used to simplify terms involving the magnitudes of $\vec{a}$ and $\vec{b}$.

Step-by-Step Solution

Step 1: Simplify the given equation using Componendo and Dividendo. The given equation is: $\frac{|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|} = \sqrt{2} + 1$ Let $P = |\vec{a} + \vec{b}|$ and $Q = |\vec{a} - \vec{b}|$. The equation becomes $\frac{P+Q}{P-Q} = \sqrt{2} + 1$. We apply the Componendo and Dividendo rule to the left side, which means we will form a new ratio by adding the numerator and denominator, and then subtracting the denominator from the numerator. We apply the same operation to the right side. $\frac{(|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|) + (|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|)}{(|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|) - (|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|)} = \frac{(\sqrt{2} + 1) + 1}{(\sqrt{2} + 1) - 1}$ Simplify the numerator and denominator on both sides: $\frac{2|\vec{a} + \vec{b}|}{2|\vec{a} - \vec{b}|} = \frac{\sqrt{2} + 2}{\sqrt{2}}$ $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = \frac{\sqrt{2}(1 + \sqrt{2})}{\sqrt{2}}$ $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = 1 + \sqrt{2}$ This gives us a simpler relationship between the magnitudes of the sum and difference of the vectors.

Step 2: Square both sides and substitute magnitude formulas. We square both sides of the simplified equation from Step 1: $\left(\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|}\right)^2 = (1 + \sqrt{2})^2$ $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a} - \vec{b}|^2} = 1^2 + (\sqrt{2})^2 + 2(1)(\sqrt{2})$ $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a} - \vec{b}|^2} = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$ Now, we use the formulas for the square of vector magnitudes and the given condition $|\vec{a}| = |\vec{b}|$. $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} = 2|\vec{a}|^2 + 2\vec{a} \cdot \vec{b}$. $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} = 2|\vec{a}|^2 - 2\vec{a} \cdot \vec{b}$. Substitute these into the equation from the previous step: $\frac{2|\vec{a}|^2 + 2\vec{a} \cdot \vec{b}}{2|\vec{a}|^2 - 2\vec{a} \cdot \vec{b}} = 3 + 2\sqrt{2}$ Factor out 2 from the numerator and denominator on the left side: $\frac{2(|\vec{a}|^2 + \vec{a} \cdot \vec{b})}{2(|\vec{a}|^2 - \vec{a} \cdot \vec{b})} = 3 + 2\sqrt{2}$ $\frac{|\vec{a}|^2 + \vec{a} \cdot \vec{b}}{|\vec{a}|^2 - \vec{a} \cdot \vec{b}} = 3 + 2\sqrt{2}$

Step 3: Solve for the ratio $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$. Let $x = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$. We can rewrite the equation from Step 2 by dividing the numerator and denominator by $|\vec{a}|^2$: $\frac{\frac{|\vec{a}|^2}{|\vec{a}|^2} + \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}}{\frac{|\vec{a}|^2}{|\vec{a}|^2} - \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}} = 3 + 2\sqrt{2}$ $\frac{1 + x}{1 - x} = 3 + 2\sqrt{2}$ Now, we solve for $x$: $1 + x = (3 + 2\sqrt{2})(1 - x)$ $1 + x = 3 + 2\sqrt{2} - (3 + 2\sqrt{2})x$ $x + (3 + 2\sqrt{2})x = 3 + 2\sqrt{2} - 1$ $x(1 + 3 + 2\sqrt{2}) = 2 + 2\sqrt{2}$ $x(4 + 2\sqrt{2}) = 2 + 2\sqrt{2}$ $x = \frac{2 + 2\sqrt{2}}{4 + 2\sqrt{2}}$ We can simplify this expression for $x$ by dividing the numerator and denominator by 2: $x = \frac{1 + \sqrt{2}}{2 + \sqrt{2}}$ To rationalize the denominator, multiply the numerator and denominator by the conjugate of $2 + \sqrt{2}$, which is $2 - \sqrt{2}$: $x = \frac{(1 + \sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}$ $x = \frac{1(2) + 1(-\sqrt{2}) + \sqrt{2}(2) + \sqrt{2}(-\sqrt{2})}{2^2 - (\sqrt{2})^2}$ $x = \frac{2 - \sqrt{2} + 2\sqrt{2} - 2}{4 - 2}$ $x = \frac{\sqrt{2}}{2}$ So, we have found that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} = \frac{\sqrt{2}}{2}$.

Step 4: Calculate the required expression $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$. We need to find $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$. From Step 2, we have: $|\vec{a} + \vec{b}|^2 = 2|\vec{a}|^2 + 2\vec{a} \cdot \vec{b}$ Divide both sides by $|\vec{a}|^2$: $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = \frac{2|\vec{a}|^2}{|\vec{a}|^2} + \frac{2\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$ $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = 2 + 2\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)$ Substitute the value of $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$ that we found in Step 3: $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = 2 + 2\left(\frac{\sqrt{2}}{2}\right)$ $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = 2 + \sqrt{2}$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with square roots and fractions. Double-check expansions and simplifications.
• Misapplication of Componendo and Dividendo: Ensure the rule is applied correctly to the ratio of magnitudes, not directly to the vectors themselves.
• Forgetting $|\vec{a}| = |\vec{b}|$: This condition is crucial for simplifying $|\vec{a}|^2 + |\vec{b}|^2$ to $2|\vec{a}|^2$, which is a key step in solving the problem.

Summary The problem was solved by first simplifying the given complex ratio of vector magnitudes using the Componendo and Dividendo rule. This led to a simpler relationship between $|\vec{a} + \vec{b}|$ and $|\vec{a} - \vec{b}|$. Squaring this relationship and substituting the vector magnitude expansion formulas, along with the given condition $|\vec{a}| = |\vec{b}|$, allowed us to form an equation involving the ratio $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$. Solving for this ratio and then substituting it back into the expression for $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$ yielded the final answer.

The final answer is \boxed{2 + \sqrt{2}}. which corresponds to option (A).