The correct option is A 5^i+^j+2^k
Let →b=x^i+y^j+z^k
Given →a⋅→b=14⇒2x+2y+z=14(1)
→a×→b=3^i+^j−8^k
→a×→b=∣∣
∣
∣∣^i^j^k221xyz∣∣
∣
∣∣=3^i+^j−8^k
2z−y=3⇒y=2z−3(2)
−(2z−x)=1⇒x=2z+1(3)
Substitute equation (2),(3) in equation (1), we get
z=2, y=1, x=5
∴→b=5^i+^j+2^k