Vectors & 3D GeometryeasyPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Max of 3|3a+2b| + 4|3a−2b| with |a|=2, |b|=3: 60 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
If a⃗\vec a and b⃗\vec b are two vectors such that ∣a⃗∣=2|\vec a|=2 and ∣b⃗∣=3|\vec b|=3, then the maximum value of 3∣(3a⃗+2b⃗)∣+4∣(3a⃗−2b⃗)∣3\big|(3\vec a+2\vec b)\big|+4\big|(3\vec a-2\vec b)\big| is
A3030
B3636
C6060correct
D7272
Solution
Step 1: Let θ\theta be the angle between a⃗\vec a and b⃗\vec b, so a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=6cos⁡θ\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=6\cos\theta. Then
∣3a⃗+2b⃗∣2=9(4)+4(9)+12a⃗⋅b⃗=72+72cos⁡θ,|3\vec a+2\vec b|^2=9(4)+4(9)+12\vec a\cdot\vec b=72+72\cos\theta,
∣3a⃗−2b⃗∣2=72−72cos⁡θ.|3\vec a-2\vec b|^2=72-72\cos\theta.
 Step 2: So
E=372+72cos⁡θ+472−72cos⁡θ=3144cos⁡2θ2+4144sin⁡2θ2,E=3\sqrt{72+72\cos\theta}+4\sqrt{72-72\cos\theta}=3\sqrt{144\cos^2\tfrac\theta2}+4\sqrt{144\sin^2\tfrac\theta2},
 using 1+cos⁡θ=2cos⁡2θ21+\cos\theta=2\cos^2\tfrac\theta2 and 1−cos⁡θ=2sin⁡2θ21-\cos\theta=2\sin^2\tfrac\theta2 (and 72⋅2=14472\cdot2=144):
E=36cos⁡θ2+48sin⁡θ2.E=36\cos\frac\theta2+48\sin\frac\theta2.
 Step 3: The maximum of Acos⁡ϕ+Bsin⁡ϕA\cos\phi+B\sin\phi is A2+B2\sqrt{A^2+B^2}:
Emax⁡=362+482=1296+2304=3600=60.E_{\max}=\sqrt{36^2+48^2}=\sqrt{1296+2304}=\sqrt{3600}=60.
 Correct answer: (3)
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