1. Given vectors $\vec a=\hat i+\hat j=(1,1,0),\qquad \vec b=3\hat j+4\hat k=(0,3,4)$

2. Resolve $\vec b$ into components parallel and perpendicular to $\vec a$

   Let $\vec b=\vec b_1+\vec b_2$ where $\vec b_1 \parallel \vec a$ and $\vec b_2 \perp \vec a$.

   The component of $\vec b$ parallel to $\vec a$ is the projection of $\vec b$ on $\vec a$: $\vec b_1=\operatorname{proj}_{\vec a}\vec b=\frac{\vec b\cdot \vec a}{|\vec a|^2}\vec a$

   Compute: $\vec b\cdot \vec a=(0)(1)+(3)(1)+(4)(0)=3$ $|\vec a|^2=1^2+1^2=2$

   Hence, $\vec b_1=\frac{3}{2}(\hat i+\hat j)=\frac{3}{2}\hat i+\frac{3}{2}\hat j$

3. Find $\vec b_2$ $\vec b_2=\vec b-\vec b_1$ $\vec b_2=(0,3,4)-\left(\frac32,\frac32,0\right)=\left(-\frac32,\frac32,4\right)$

   So, $\vec b_2=-\frac32\hat i+\frac32\hat j+4\hat k$

4. Compute $\vec b_1\times \vec b_2$

   \begin{vmatrix} \hat i & \hat j & \hat k\\ \frac32 & \frac32 & 0\\ -\frac32 & \frac32 & 4 \end{vmatrix}$$ Expanding: $$=\hat i\left(\frac32\cdot 4-0\cdot \frac32\right) -\hat j\left(\frac32\cdot 4-0\cdot \left(-\frac32\right)\right) +\hat k\left(\frac32\cdot \frac32-\frac32\cdot \left(-\frac32\right)\right)$$ $$=6\hat i-6\hat j+\hat k\left(\frac94+\frac94\right)$$ $$=6\hat i-6\hat j+\frac92\hat k$$

5. Match with options This is exactly Option B.

$\boxed{\vec b_1\times \vec b_2=6\hat i-6\hat j+\frac92\hat k}$