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 :

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Vector Algebra

Let

a

and

b

be the vectors of the same magnitude such that

\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1

. Then

\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}}

is :

Answer: (C)

With

∣ a ∣ = ∣ b ∣

, apply componendo and dividendo to

\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1

to obtain

\frac{∣ a + b ∣}{∣ a - b ∣} = \frac{2 + 2}{2} = 1 + 2

. Squaring,

\frac{∣ a + b ∣ ^{2}}{∣ a - b ∣ ^{2}} = 3 + 22

. Since

∣ a + b ∣^{2} + ∣ a - b ∣^{2} = 2∣ a ∣^{2} + 2∣ b ∣^{2} = 4∣ a ∣^{2}

, one gets

∣ a - b ∣^{2} = \frac{4∣ a ∣ ^{2}}{4 + 2 2}

, and then

∣ a + b ∣^{2} = (3 + 22) \frac{4∣ a ∣ ^{2}}{4 + 2 2} = (2 + 2) ∣ a ∣^{2}

. Hence

\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}} = 2 + 2

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option