Answer
\[
\hat{k}-\hat{\jmath}=\hat{i} \times(\hat{\imath}+\hat{\jmath}+\hat{k})
\]
Work Step by Step
For two 3 -Dimensional vectors
\[
\vec{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle \text { and } \quad \vec{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle \quad
\]
its 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) Since $\hat{i}=\langle 1,0,0\rangle$ and $\hat{\imath}+\hat{\jmath}+\hat{k}=\langle 1,1,1\rangle$, we get that
\[
\begin{aligned}
\hat{i} \times(\hat{i}+\hat{j}+\hat{k}) &=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 0 \\
1 & 1 & 1
\end{array}\right| \\
&=\hat{k}-\hat{j}
\end{aligned}
\]
b) Notice that
\[
\begin{aligned}
(\hat{k}+\hat{j}+\hat{i}) \times \hat{i} &=\hat{i} \times \hat{k} +\hat{i} \times \hat{j}+\hat{i} \times \hat{i}\\
&=-(\hat{k} \times \hat{i}) +0+\hat{k} \\
&=\hat{k}-\hat{j}
\end{aligned}
\]
We get that:
\[
\hat{k}-\hat{\jmath}=\hat{i} \times(\hat{i}+\hat{\jmath}+\hat{k})
\]
