We have, (a+3b)⊥(7a−5b)
Therefore, (a+3b)⋅(7a−5b)=0
⇒7∣a∣2−15∣b∣2+16a⋅b=0…(1)
and (a−4b)⋅(7a−2b)=0
⇒7∣a∣2+8∣b∣2−30a⋅b=0…(2)
From (1) and (2), we get
2∣a∣cosθ=∣b∣
∴cosθ=2∣a∣∣b∣
⇒θ=60∘.
JEE Main 2021 — Mathematics Vectors & 3D Geometry
If a and b are unit vectors and (a+3b) is perpendicular to (7a−5b) and (a−4b) is perpendicular to (7a−2b), then the angle between a and b (in degrees) is _________.
Held on 25 Jul 2021 · Verified 6 Jul 2026.
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