The correct option is A 1√5(−^j−2^k)
We know that the unit vector perpendicular to two vectors is given as
^n=→A×→B|→A×→B|
So, we have →A×→B as per determinant method as
⎡⎢⎣^i^j^k12−13−21⎤⎥⎦
⇒ (2−2)^i−(1−(−3))^j+(−2−6)^k
⇒ −4^j−8^k
Also, |→A×→B|=√(−4)2+(−8)2=√80=4√5
Hence, ^n=→A×→B|→A×→B|=−4^j−8^k4√5=1√5(−^j−2^k)