We know the identity |a×b|2+(a⋅b)2=|a|2|b|2. Since |a|=7 and |b|=1, the right-hand side is 72⋅12=49 The problem states |a×b|2=k2−(a⋅b)2. Comparing with the identity, we get k2=49‌‌⟹‌‌k=7. No further restriction on the angle θ is needed-this relation holds for any θ. Answer: k=7, and θ can be arbitrary. (Option D )