Question Number 150405 by liberty last updated on 12/Aug/21
Answered by EDWIN88 last updated on 12/Aug/21
Answered by ajfour last updated on 12/Aug/21
Commented by ajfour last updated on 12/Aug/21
$$\mathrm{4}{cm}={a} \\ $$$$\overline {{AE}}={a}\left({i}+{j}\right) \\ $$$${AH}:\:\:\:{r}=\lambda\left({i}+{j}+{k}\right) \\ $$$${ME}:\:{r}={a}\left({i}+{j}\right)+\mu{a}\left({i}+\frac{{j}}{\mathrm{2}}{j}−\frac{{k}}{\mathrm{2}}\right) \\ $$$${s}=\hat {{n}}=\frac{\overline {{AH}}×\overline {{ME}}}{\mid\overline {{AH}}×\overline {{ME}}\mid} \\ $$$$\:\:\:\:=\:\frac{\begin{vmatrix}{{i}}&{{j}}&{{k}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{1}}\end{vmatrix}}{\bigtriangleup}=\frac{−\mathrm{2}{i}+\mathrm{3}{j}−{k}}{\:\sqrt{\mathrm{14}}} \\ $$$${s}={perpendicular}\:{distance}\:{b}/{w} \\ $$$$\:\:\:\:\:{AH}\:{and}\:{EM}={projection} \\ $$$${of}\:{AE}\:\:{along}\:\hat {{n}} \\ $$$$\:\:\:=\:\overline {{AE}}\centerdot\hat {{n}} \\ $$$$\:\:\:=\:{a}\left({i}+{j}\right)\centerdot\left(\frac{−\mathrm{2}{i}+\mathrm{3}{j}−{k}}{\:\sqrt{\mathrm{14}}}\right) \\ $$$$\:\:{s}=\frac{{a}}{\:\sqrt{\mathrm{14}}}=\frac{\mathrm{4}{cm}}{\:\sqrt{\mathrm{14}}}=\frac{\mathrm{2}\sqrt{\mathrm{14}}}{\mathrm{7}}{cm}. \\ $$
Commented by EDWIN88 last updated on 12/Aug/21
