If a and b are two vectors such that | a | = 2 and | b | = 3 , then the maximum value of 3 | (3 a + 2 b )| + 4|(3 a - 2 b )| is :

Correct answer: C. 60 Let θ be the angle between a and b . We have |3 a + 2 b | = 9| a |^2 + 4| b |^2 + 12| a || b | θ Substituting | a | = 2 and | b | = 3 : |3 a + 2 b | = 9(4) + 4(9) + 12(2)(3) θ = 36 + 36 + 72 θ |3 a + 2 b | = 72(1+ θ) = 144 ^2((θ)/(2)) = 12 ((θ)/(2)) Similarly, |3 a - 2 b | = 9| a |^2 + 4| b |^2 - 12| a || b | θ |3 a - 2 b | = 36 + 36 - 72 θ = 72(1- θ) = 144 ^2((θ)/(2)) = 12 ((θ)/(2)) The given expression becomes: 3|3 a + 2 b | + 4|3 a - 2 b | = 3(12 ((θ)/(2))) + 4(12 ((θ)/(2))) = 36 ((θ)/(2)) + 48 ((θ)/(2)) The maximum value of A x + B x is A^2 + B^2 . Maximum value = 36^2 + 48^2 = 1296 + 2304 = 3600 = 60 Answer: 60

JEE Main 20262 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 2 April · Morning Shift · Mathematics · Vector Algebra

If a and b are two vectors such that | a | = 2 and | b | = 3 , then...

\vec{a}

and

\vec{b}

are two vectors such that

|\vec{a}| = 2

and

|\vec{b}| = 3

, then the maximum value of

3

\left|

\left(3\vec{a} + 2\vec{b}\right)\right| + 4\left|\left(3\vec{a} - 2\vec{b}\right)\right|

is :

Options

A $30$
B $36$
C $60$
D $72$

Correct answer

C. $60$

Step-by-step solution

Let

\theta

be the angle between

\vec{a}

and

\vec{b}

.

We have

|3\vec{a} + 2\vec{b}| = \sqrt{9|\vec{a}|^2 + 4|\vec{b}|^2 + 12|\vec{a}||\vec{b}|\cos\theta}

Substituting

|\vec{a}| = 2

and

|\vec{b}| = 3

|3\vec{a} + 2\vec{b}| = \sqrt{9(4) + 4(9) + 12(2)(3)\cos\theta} = \sqrt{36 + 36 + 72\cos\theta}

|3\vec{a} + 2\vec{b}| = \sqrt{72(1+\cos\theta)} = \sqrt{144\cos^2\left(\dfrac{\theta}{2}\right)} = 12\cos\left(\dfrac{\theta}{2}\right)

Similarly,

|3\vec{a} - 2\vec{b}| = \sqrt{9|\vec{a}|^2 + 4|\vec{b}|^2 - 12|\vec{a}||\vec{b}|\cos\theta}

|3\vec{a} - 2\vec{b}| = \sqrt{36 + 36 - 72\cos\theta} = \sqrt{72(1-\cos\theta)} = \sqrt{144\sin^2\left(\dfrac{\theta}{2}\right)} = 12\sin\left(\dfrac{\theta}{2}\right)

The given expression becomes:

3|3\vec{a} + 2\vec{b}| + 4|3\vec{a} - 2\vec{b}| = 3\left(12\cos\left(\dfrac{\theta}{2}\right)\right) + 4\left(12\sin\left(\dfrac{\theta}{2}\right)\right)

= 36\cos\left(\dfrac{\theta}{2}\right) + 48\sin\left(\dfrac{\theta}{2}\right)

The maximum value of

A\cos x + B\sin x

\sqrt{A^2 + B^2}

.

Maximum value

= \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60

Answer:

60

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