Given a=a1i^+a2j^+a3k^
Since the vector makes equal angles with the coordinate axes OX,OY and OZ, soa=λ(31i^+31j^+31k^)=3λ(i^+j^+k^)
Now projection of aon b=7
⇒3λ5(i^+j^+k^)⋅(3i^+4j^)=7
⇒λ=53
So a=5(i^^+j^+k^)
Let b=5α(i^+j^+k^)+β(i^)
Since both vectors are perpendicular a⋅b=0
⇒25α(3)+5β=0
⇒15α+β=0⇒β=−15α
i.e. b=5α(−2i^+j^+k^)
∣b∣=53
b=±25(−2i^+j^+k^)
Projection of b on 3i^+4j^ is
5b⋅(3i^+4j^)=±25(5−6+4)=±2