Let a and b be the vectors of the same magnitude such that |a + b| + |a - b| / |a + b| - |a - b| = sqrt(2) + 1. Then |a + b|^2 / |a|^2 is :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q15 – JEE-MAIN 2025

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Home Maths Vector AlgebraQuestion #523

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.

Then

\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}

is :

Clue 1 of 3: Vector Magnitudes

Recall the formulas for the magnitudes of the sum and difference of two vectors, especially when their magnitudes are equal.

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Step 1: Simplify the given ratio using Componendo and Dividendo✦ Active

Let $| \vec{a} | = | \vec{b} | = k$ . Let $X = | \vec{a} + \vec{b} |$ and $Y = | \vec{a} - \vec{b} |$ . The given equation is $\frac{X + Y}{X - Y} = \sqrt{2} + 1$ . Applying componendo and dividendo:

\frac{(X + Y) + (X - Y)}{(X + Y) - (X - Y)} = \frac{(\sqrt{2} + 1) + 1}{(\sqrt{2} + 1) - 1}

\frac{2 X}{2 Y} = \frac{\sqrt{2} + 2}{\sqrt{2}}

\frac{X}{Y} = \frac{\sqrt{2} (1 + \sqrt{2})}{\sqrt{2}} = 1 + \sqrt{2}

So, $\frac{| \vec{a} + \vec{b} |}{| \vec{a} - \vec{b} |} = 1 + \sqrt{2}$ .

Step 2: Express magnitudes in terms of dot product and find

\cos θ

○ Expand

Square both sides of the simplified ratio:

\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} - \vec{b} |^{2}} = (1 + \sqrt{2})^{2} = 1 + 2 + 2 \sqrt{2} = 3 + 2 \sqrt{2}

We know $| \vec{a} + \vec{b} |^{2} = | \vec{a} |^{2} + | \vec{b} |^{2} + 2 \vec{a} \cdot \vec{b} = 2 k^{2} + 2 k^{2} \cos θ$ and $| \vec{a} - \vec{b} |^{2} = | \vec{a} |^{2} + | \vec{b} |^{2} - 2 \vec{a} \cdot \vec{b} = 2 k^{2} - 2 k^{2} \cos θ$ . Substituting these into the squared ratio:

\frac{2 k^{2} (1 + \cos θ)}{2 k^{2} (1 - \cos θ)} = \frac{1 + \cos θ}{1 - \cos θ} = 3 + 2 \sqrt{2}

Solving for $\cos θ$ :

1 + \cos θ = (3 + 2 \sqrt{2}) (1 - \cos θ)

1 + \cos θ = 3 - 3 \cos θ + 2 \sqrt{2} - 2 \sqrt{2} \cos θ

4 \cos θ + 2 \sqrt{2} \cos θ = 2 + 2 \sqrt{2}

\cos θ (4 + 2 \sqrt{2}) = 2 (1 + \sqrt{2})

\cos θ = \frac{2 (1 + \sqrt{2})}{2 (2 + \sqrt{2})} = \frac{1 + \sqrt{2}}{2 + \sqrt{2}} = \frac{(1 + \sqrt{2}) (2 - \sqrt{2})}{(2 + \sqrt{2}) (2 - \sqrt{2})} = \frac{2 - \sqrt{2} + 2 \sqrt{2} - 2}{4 - 2} = \frac{\sqrt{2}}{2}

Step 3: Calculate the required expression○ Expand

We need to find $\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}$ :

\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}} = \frac{2 k^{2} + 2 \vec{a} \cdot \vec{b}}{k^{2}} = \frac{2 k^{2} + 2 k^{2} \cos θ}{k^{2}} = 2 (1 + \cos θ)

Substitute the value of $\cos θ = \frac{\sqrt{2}}{2}$ :

2 (1 + \frac{\sqrt{2}}{2}) = 2 + \sqrt{2}

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Maths Question 15 – JEE-MAIN 2025

Clue 1 of 3: Vector Magnitudes

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