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

Correct answer: C. 2+ 2 | a + b |+| a - b | | a + b |-| a - b | = 2 +1 Apply componendo and dividendo aligned & 2| a + b | 2| a - b | = 2 +2 2 & | a + b |=(1+ 2 )| a - b | & | a + b |^2=(3+2 2 )| a - b |^2 & 2| a |^2+2 a · ~b =(3+2 2 )(2| a |^2-2 a · ~b ) & 2| a |^2(2+2 2 )=2 a · ~b (4+2 2 ) & a · ~b | a |^2 = 2+2 2 4+2 2 = 1 2 aligned Now aligned & | a + b |^2 | a |^2 =1+ | b |^2 | a |^2 + 2 a · ~b | a |^2 & =1+1+2( 1 2 )=2+ 2 aligned

JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Vector Algebra

Let a and b be the vectors of the same magnitude such that | a + b...

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 :

Options

A $2+4 \sqrt{2}$
B $1+\sqrt{2}$
C $2+\sqrt{2}$
D $4+2 \sqrt{2}$

Correct answer

C. $2+\sqrt{2}$

Step-by-step solution

\frac{|\bar{a}+\bar{b}|+|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|-|\bar{a}-\bar{b}|}=\sqrt{2}+1

Apply componendo and dividendo

\begin{aligned}<br/>& \Rightarrow \frac{2|\overline{\mathrm{a}}+\overline{\mathrm{b}}|}{2|\overline{\mathrm{a}}-\overline{\mathrm{b}}|}=\frac{\sqrt{2}+2}{\sqrt{2}} \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=(1+\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}| \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=(3+2 \sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2 \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=(3+2 \sqrt{2})\left(2|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}\right) \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2(2+2 \sqrt{2})=2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}(4+2 \sqrt{2}) \\ & \Rightarrow \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2}=\frac{2+2 \sqrt{2}}{4+2 \sqrt{2}}=\frac{1}{\sqrt{2}}<br/>\end{aligned}

Now

\begin{aligned}<br/>& \frac{|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}=1+\frac{|\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}+\frac{2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2} \\ & =1+1+2\left(\frac{1}{\sqrt{2}}\right)=2+\sqrt{2}<br/>\end{aligned}

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