Question

Let $ \vec{a} $ and $ \vec{b} $ be the vectors of the same magnitude such that $ \frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1. \quad \text{Then } \frac{| \vec{a} + \vec{b} |^2}{| \vec{a} |^2} \text{ is:} $

• A. \( 2 + 4\sqrt{2} \)
• B. \( 1 + \sqrt{2} \)
• C. \( 2 + \sqrt{2} \)
• D. \( 4 + 2\sqrt{2} \)

Hint:

When dealing with vector magnitudes, use trigonometric identities to simplify the expressions for \( | \vec{a} + \vec{b} | \) and \( | \vec{a} - \vec{b} | \).

Answer 1

The Correct Option is C

To address the problem, we commence by analyzing the provided condition:

\[\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1\]

Given that vectors \(\vec{a}\) and \(\vec{b}\) share the same magnitude, we have \(|\vec{a}| = |\vec{b}| = a\)\). Utilizing the identity for magnitudes:

\[|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}} = \sqrt{2a^2 + 2a^2\cos\theta}\]\[|\vec{a} - \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}} = \sqrt{2a^2 - 2a^2\cos\theta}\]

Let us define:

• \(x = |\vec{a} + \vec{b}|\)\)
• \(y = |\vec{a} - \vec{b}|\)\)

Then, applying the identity yields:

\[x + y = \sqrt{2a^2(1 + \cos\theta)} + \sqrt{2a^2(1 - \cos\theta)} \] \] \] \] \] \] \[ x - y = \sqrt{2a^2(1 + \cos\theta)} - \sqrt{2a^2(1 - \cos\theta)} \] \] \] \] \]\]

Now, we re-examine the initial condition:

\[\frac{x + y}{x - y} = \sqrt{2} + 1 \] \] \] \] \] \] \[ \Rightarrow x + y = (\sqrt{2} + 1)(x - y) \] \] \] \] \]\]

By substituting \(x = ka\)\) and \(y = ma\)\), where \(k\)\) and \(m\)\) are undetermined constants, we solve for:

\[k + m = (\sqrt{2} + 1)(k - m) \] \] \] \] \] \] \[ \Rightarrow k + m = (\sqrt{2} + 1)(k - m) \] \] \] \] \]\]

Cross-multiplication and simplification lead to \(\cos\theta = \frac{1}{\sqrt{2}}\). Consequently, we can determine \(|\vec{a} + \vec{b}|^2 = 2a^2 + 2a^2\frac{1}{\sqrt{2}}\)\):

\[\Rightarrow |\vec{a} + \vec{b}|^2 = 2a^2 + \sqrt{2}a^2\]

Therefore,

\[\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = 2 + \sqrt{2}\]

The expression evaluates to: \(2 + \sqrt{2}\).