If the vector ⃗b=3^j+4^k is written as the sum of a vector ⃗b1, parallel to ⃗a=^i+^j and a vector ⃗b2, perpendicular to ⃗a, then ⃗b1 × ⃗b2 is equal to.

The correct option is C
6^i−6^j+92^k
Given, ⃗a=^i+^j
Since, ⃗b1 is parallel to ⃗a, we can say ⃗b1=k(^i+^j)
Let ⃗b2=p^i+q^j+r^k
Since ⃗a ⊥ ⃗b2, ⃗a.⃗b2=0
⇒(p^i+q^j+r^k)⋅(^i+^j)=0
⇒p+q=0
Since, ⃗b=⃗b1+⃗b2
∴3^j+4^k=k(^i+^j)+(p^i−p^j+r^k)
Comparing the components, we get
0=k+p(1)
3=k−p(2)
4=r(3)
Adding (1) and (2), 2k=3 ⇒k=3/2
From (1), p=−k=−3/2
Hence, ⃗b1=3/2^i+3/2^j and ⃗b2=−3/2^i+3/2^j+4^k
⇒⃗b1 × ⃗b2=∣∣∣∣∣∣^i^j^k3/23/20−3/23/24∣∣∣∣∣∣
=^i(6)−^j(6)+^k(9/4+9/4)=6^i−6^j+92^k