Question Number 120801 by TITA last updated on 02/Nov/20 | ||
Commented by liberty last updated on 02/Nov/20 | ||
$$\mathrm{let}\:\begin{cases}{\mathrm{i}=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)}\\{\mathrm{j}=\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)}\\{\mathrm{k}=\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)}\end{cases}\rightarrow\begin{cases}{\mathrm{i}.\mathrm{i}=\mathrm{1}.\mathrm{1}+\mathrm{0}.\mathrm{0}+\mathrm{0}.\mathrm{0}=\mathrm{1}}\\{\mathrm{j}.\mathrm{j}=\mathrm{0}.\mathrm{0}+\mathrm{1}.\mathrm{1}+\mathrm{0}.\mathrm{0}=\mathrm{1}}\\{\mathrm{k}.\mathrm{k}=\mathrm{0}.\mathrm{0}+\mathrm{0}.\mathrm{0}+\mathrm{1}.\mathrm{1}=\mathrm{1}}\end{cases} \\ $$ | ||
Commented by TITA last updated on 03/Nov/20 | ||
$${thanks} \\ $$ | ||
Answered by $@y@m last updated on 03/Nov/20 | ||