The vector (a→+3b→) is perpendicular to (7a→−5b→) and (a→−4b→) is perpendicular to (7a→−2b→). Find the angle between a→ and b→ in degrees.

(a→+3b→)⋅(7a→−5b→)=0⇒7a2−15b2+16a→⋅b→=0  ........(i) and (a→−4b→)⋅(7a→−2b→)=0⇒7a2+8b2−30a→⋅b→=0  .........(ii)By subtracting (i) and (ii)⇒−23b2+46a→⋅b→=0⇒2a→⋅b→=b2 So 7a2−15b2+8b2=0⇒a2=b2⇒2abcos⁡θ=b2⇒2cos⁡θ=1⇒θ=cos−1⁡(1/2)=60∘