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

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2026

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Home Maths Vector AlgebraQuestion #525

\vec{a}

and

\vec{b}

are two vectors such that

| \vec{a} | = 2

and

| \vec{b} | = 3

, then the maximum value of

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

is:

Clue 1 of 3: Vector Magnitudes and Parallelogram Law

The problem involves magnitudes of sums and differences of vectors. Recall the parallelogram law relating these magnitudes.

🥷

Ninja Strategy•Parallelogram Law and Cauchy-Schwarz

Use the parallelogram law to establish a relationship between the squares of the magnitudes, then apply the Cauchy-Schwarz inequality to find the maximum value of the linear combination.

🧠 Full Solution Path

Step 1: Define new vectors and their magnitudes✦ Active

Let $\vec{X} = 3 \vec{a}$ and $\vec{Y} = 2 \vec{b}$ . Given $| \vec{a} | = 2$ and $| \vec{b} | = 3$ . We can find the magnitudes of $\vec{X}$ and $\vec{Y}$ :

| \vec{X} | = | 3 \vec{a} | = 3 | \vec{a} | = 3 (2) = 6

| \vec{Y} | = | 2 \vec{b} | = 2 | \vec{b} | = 2 (3) = 6

The expression to maximize is $E = | 3 (\vec{3 a} + \vec{2 b}) | + | 4 (\vec{3 a} - \vec{2 b}) |$ , which simplifies to $E = 3 | \vec{3 a} + \vec{2 b} | + 4 | \vec{3 a} - \vec{2 b} |$ . Substituting $\vec{X}$ and $\vec{Y}$ :

E = 3 | \vec{X} + \vec{Y} | + 4 | \vec{X} - \vec{Y} |

Step 2: Apply the Parallelogram Law○ Expand

The parallelogram law states that for any two vectors $\vec{X}$ and $\vec{Y}$ :

| \vec{X} + \vec{Y} |^{2} + | \vec{X} - \vec{Y} |^{2} = 2 (| \vec{X} |^{2} + | \vec{Y} |^{2})

Substitute the magnitudes of $\vec{X}$ and $\vec{Y}$ :

| \vec{X} + \vec{Y} |^{2} + | \vec{X} - \vec{Y} |^{2} = 2 (6^{2} + 6^{2}) = 2 (36 + 36) = 2 (72) = 144

Let $P = | \vec{X} + \vec{Y} |$ and $Q = | \vec{X} - \vec{Y} |$ . Then we have the constraint:

P^{2} + Q^{2} = 144

💡 Teacher's Secret Hint

Remember that P and Q are magnitudes, so they must be non-negative.

Step 3: Maximize the expression using Cauchy-Schwarz Inequality○ Expand

We need to maximize $E = 3 P + 4 Q$ subject to $P^{2} + Q^{2} = 144$ . We can use the Cauchy-Schwarz inequality, which states that for real numbers $a_{1}, a_{2}, b_{1}, b_{2}$ :

(a_{1} b_{1} + a_{2} b_{2})^{2} \leq (a_{1}^{2} + a_{2}^{2}) (b_{1}^{2} + b_{2}^{2})

Applying this to our expression with $(a_{1}, a_{2}) = (3, 4)$ and $(b_{1}, b_{2}) = (P, Q)$ :

(3 P + 4 Q)^{2} \leq (3^{2} + 4^{2}) (P^{2} + Q^{2})

Substitute the known values:

(3 P + 4 Q)^{2} \leq (9 + 16) (144)

(3 P + 4 Q)^{2} \leq (25) (144)

Taking the square root of both sides (since $P, Q$ are magnitudes, $3 P + 4 Q$ must be non-negative):

3 P + 4 Q \leq \sqrt{25 \times 144}

3 P + 4 Q \leq 5 \times 12

3 P + 4 Q \leq 60

The maximum value of the expression is 60. This maximum is achievable when $(P, Q)$ is proportional to $(3, 4)$ , which corresponds to a specific angle between $\vec{a}$ and $\vec{b}$ .

💡 Teacher's Secret Hint

The equality in Cauchy-Schwarz holds when the vectors $(a_{1}, a_{2})$ and $(b_{1}, b_{2})$ are proportional.

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Maths Question 14 – JEE-MAIN 2026

Clue 1 of 3: Vector Magnitudes and Parallelogram Law

🧠 Full Solution Path