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:

Given: | a| = | b| and | a + b | + | a - b || a + b | - | a - b | = 2 + 1 Find: | a + b |^2| a |^2 Let | a| = | b| = a and let the angle between them be. Then | a + b| = | a|^2 + | b|^2 + 2 a b = 2a^2 + 2a^2 and | a - b| = | a|^2 + | b|^2 - 2 a b = 2a^2 - 2a^2 Set x = | a + b|, y = | a - b| Then the given condition becomes x+yx-y = 2+1 so x+y = ( 2+1)(x-y) Using the working stated in the solution, this gives = 1 2 Hence | a + b|^2 = 2a^2 + 2a^2 = 2a^2 + 2a^2 1 2 = 2a^2 + 2a^2 Therefore, | a + b|^2| a|^2 = 2a^2 + 2a^2a^2 = 2 + 2 Therefore, the correct option is C. Given: | a| = | b| = a. Find: | a + b |^2| a|^2 Write the magnitudes as | a + b| = 2a^2(1+), | a - b| = 2a^2(1-) Now use 1+ = 2 ^2 2, 1- = 2 ^2 2 So | a + b| = 2a 2, | a - b| = 2a 2 Substituting into the ratio, 2 + 2 2 - 2 = 2+1 This is satisfied when 2 = 2-1 which gives = 45^ = 1 2 Now | a+ b|^2| a|^2 = 2a^2(1+)a^2 = 2 (1+ 1 2) = 2+ 2 Hence, the required value is 2+ 2, so the correct option is C.

Let a and b be the vectors of the same magnitude such that | a +… — Solution | JEE Mathematics

MCQMediumJEE 2025Basics of Vectors

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:

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

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $|\vec{a}| = |\vec{b}|$ and

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

Find: $\frac{| \vec{a} + \vec{b} |^2}{| \vec{a} |^2}$

Let $|\vec{a}| = |\vec{b}| = a$ and let the angle between them be $\theta$ . Then

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

and

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

Set

x = |\vec{a} + \vec{b}|, \qquad y = |\vec{a} - \vec{b}|

Then the given condition becomes

\frac{x+y}{x-y} = \sqrt{2}+1

x+y = (\sqrt{2}+1)(x-y)

Using the working stated in the solution, this gives

\cos\theta = \frac{1}{\sqrt{2}}

Hence

|\vec{a} + \vec{b}|^2 = 2a^2 + 2a^2\cos\theta = 2a^2 + 2a^2\cdot\frac{1}{\sqrt{2}} = 2a^2 + \sqrt{2}a^2

Therefore,

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

Therefore, the correct option is C.

Using half-angle identities

Given: $|\vec{a}| = |\vec{b}| = a$ . Find: $\frac{| \vec{a} + \vec{b} |^2}{|\vec{a}|^2}$

Write the magnitudes as

|\vec{a} + \vec{b}| = \sqrt{2a^2(1+\cos\theta)}, \qquad |\vec{a} - \vec{b}| = \sqrt{2a^2(1-\cos\theta)}

Now use

1+\cos\theta = 2\cos^2\frac\theta2, \qquad 1-\cos\theta = 2\sin^2\frac\theta2

|\vec{a} + \vec{b}| = 2a\cos\frac\theta2, \qquad |\vec{a} - \vec{b}| = 2a\sin\frac\theta2

Substituting into the ratio,

\frac{\cos\frac\theta2 + \sin\frac\theta2}{\cos\frac\theta2 - \sin\frac\theta2} = \sqrt{2}+1

This is satisfied when

\tan\frac\theta2 = \sqrt{2}-1

which gives

\theta = 45^\circ \quad \Rightarrow \quad \cos\theta = \frac{1}{\sqrt{2}}

Now

\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2} = \frac{2a^2(1+\cos\theta)}{a^2} = 2\left(1+\frac{1}{\sqrt{2}}\right) = 2+\sqrt{2}

Hence, the required value is $2+\sqrt{2}$ , so the correct option is C.

Common mistakes

Using $|\vec{a}+\vec{b}| = |\vec{a}| + |\vec{b}|$ directly is incorrect because magnitudes do not add linearly unless the vectors are parallel in the same direction. Use $|\vec{a} \pm \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 \pm 2\vec{a}\cdot\vec{b}$ instead.
Assuming equal magnitudes means the vectors are equal is wrong. The condition only gives $|\vec{a}| = |\vec{b}|$ , not $\vec{a} = \vec{b}$ . You must still keep the angle $\theta$ between them as an unknown.
Squaring or simplifying the ratio without first substituting suitable expressions for $|\vec{a}+\vec{b}|$ and $|\vec{a}-\vec{b}|$ can lead to algebraic errors. First rewrite both magnitudes in terms of $a$ and $\cos\theta$ , then solve the ratio carefully.

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