Find the values of $\left(4\hat{i}\right)\times (-6\hat{k})$

Let, $A=4\hat{i}$and $B=(-6\hat{k})$

Therefore, $\mathrm{A}=\hat{4}\mathrm{i}+\hat{0}\mathrm{j}+\hat{0}\mathrm{k} \mathrm{and} \mathrm{B}=\hat{0}\mathrm{i}+\hat{0}\mathrm{j}-\hat{6}\mathrm{k}$

such that,$a_1=4,a_2=0,a_3=0$ and

${\mathrm{b}}_1=0,{ \mathrm{b}}_2=0,{ \mathrm{b}}_3=-6$

Hence,

$\mathrm{A}\times \mathrm{B}=\mathrm{i}\left({\mathrm{a}}_2{ \mathrm{b}}_3-{\mathrm{b}}_2{\mathrm{a}}_3\right)\widehat{-\mathrm{j}}\left({\mathrm{a}}_1{ \mathrm{b}}_3-{\mathrm{b}}_1{\mathrm{a}}_3\right)+\hat{\mathrm{k}}\left({\mathrm{a}}_1{ \mathrm{b}}_2-{\mathrm{b}}_1{\mathrm{a}}_2\right)$

$\Rightarrow \mathrm{A}\times \mathrm{B}=\mathrm{i}(0\times -6-0\times 4)-\mathrm{j}(4\times -6-0\times 0)+\hat{k}(4\times 0-0\times 0)$

$\Rightarrow \mathrm{A}\times \mathrm{B}=\mathrm{i}\left(0\right)-\hat{j}(-24)+\hat{k}\left(0\right)$

$\Rightarrow \mathrm{A}\times \mathrm{B}=24\hat{\mathrm{j}}$