Answer
a) $(\hat{\imath}+\hat{\jmath}+\hat{k})\times\hat{\jmath} =-\hat{k}+\hat{\imath}$
b) $(\hat{\imath}+\hat{\jmath}+\hat{k})\times\hat{k} =-\hat{\imath}+\hat{\jmath}$
Work Step by Step
For two 3 -Dimensional vectors
\[
\quad \vec{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle \text { and } \vec{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle \quad
\]
The cross product is given by the determinant
\[
\vec{u} \times \vec{v}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3}
\end{array}\right|
\]
a) We need to find the cross product
\[
(\hat{i}+\hat{j}+\hat{k}) \times \hat{j}
\]
Since $\hat{j}=\langle 0,1,0\rangle$ , $\hat{i}+\hat{j}+\hat{k}=\langle 1,1,1\rangle$, we get:
\[
\begin{aligned}
(\hat{i}+\hat{j}+\hat{k}) \times\hat{j}
&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & 0 \\
1 & 1 & 1
\end{array}\right|=\hat{i}(-0+1)+\hat{j}(0-0)+\hat{k}(-1+0) \\
&=-\hat{k}+\hat{i}
\end{aligned}
\]
Notice that
\[
\begin{aligned}
(\hat{i}+\hat{j}+\hat{k})\times\hat{j} &=\hat{j} \times \hat{i}+\hat{j} \times \hat{j}+\hat{j} \times \hat{k} \\
&=-\hat{i} \times \hat{j}+\overrightarrow{0}+\hat{i} \\
&=-\hat{k}+\hat{i}=-\hat{k}+\hat{i}
\end{aligned}
\]
b) Analogously, we have that for $\hat{k}=\langle 0,0,1\rangle$:
\[
\begin{aligned}
(\hat{\imath}+\hat{\jmath}+\hat{k}) \times\hat{k}
&=\left|\begin{array}{ccc}
\hat{i} & \hat{\jmath} & \hat{k} \\
0 & 0 & 1 \\
1 & 1 & 1
\end{array}\right|=\hat{i}(0-1)-\hat{\jmath}(0-1)+\hat{k} \\
&=-\hat{i}+\hat{j}
\end{aligned}
\]
Notice that
\[
\begin{aligned}
(\hat{i}+\hat{j}+\hat{k})\times\hat{k} &=\hat{k} \times \hat{i}+\hat{k} \cdot x \hat{j}+\hat{k} \times \hat{k} \\
&=-\hat{j} \times \hat{k}+\overrightarrow{0}+\hat{j} \\
&=-\hat{i}+\hat{j}
\end{aligned}
\]
