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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

4cm=a AE^(−) =a(i+j) AH: r=λ(i+j+k) ME: r=a(i+j)+μa(i+(j/2)j−(k/2)) s=n^� =((AH^(−) ×ME^(−) )/(∣AH^(−) ×ME^(−) ∣)) = ( determinant ((i,j,k),(1,1,1),(2,1,(−1)))/△)=((−2i+3j−k)/( (√(14)))) s=perpendicular distance b/w AH and EM=projection of AE along n^� = AE^(−) ∙n^� = a(i+j)∙(((−2i+3j−k)/( (√(14))))) s=(a/( (√(14))))=((4cm)/( (√(14))))=((2(√(14)))/7)cm.

$$\mathrm{4}{cm}={a} \\ $$$$\overline {{AE}}={a}\left({i}+{j}\right) \\ $$$${AH}:\:\:\:{r}=\lambda\left({i}+{j}+{k}\right) \\ $$$${ME}:\:{r}={a}\left({i}+{j}\right)+\mu{a}\left({i}+\frac{{j}}{\mathrm{2}}{j}−\frac{{k}}{\mathrm{2}}\right) \\ $$$${s}=\hat {{n}}=\frac{\overline {{AH}}×\overline {{ME}}}{\mid\overline {{AH}}×\overline {{ME}}\mid} \\ $$$$\:\:\:\:=\:\frac{\begin{vmatrix}{{i}}&{{j}}&{{k}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{1}}\end{vmatrix}}{\bigtriangleup}=\frac{−\mathrm{2}{i}+\mathrm{3}{j}−{k}}{\:\sqrt{\mathrm{14}}} \\ $$$${s}={perpendicular}\:{distance}\:{b}/{w} \\ $$$$\:\:\:\:\:{AH}\:{and}\:{EM}={projection} \\ $$$${of}\:{AE}\:\:{along}\:\hat {{n}} \\ $$$$\:\:\:=\:\overline {{AE}}\centerdot\hat {{n}} \\ $$$$\:\:\:=\:{a}\left({i}+{j}\right)\centerdot\left(\frac{−\mathrm{2}{i}+\mathrm{3}{j}−{k}}{\:\sqrt{\mathrm{14}}}\right) \\ $$$$\:\:{s}=\frac{{a}}{\:\sqrt{\mathrm{14}}}=\frac{\mathrm{4}{cm}}{\:\sqrt{\mathrm{14}}}=\frac{\mathrm{2}\sqrt{\mathrm{14}}}{\mathrm{7}}{cm}. \\ $$

Commented by EDWIN88 last updated on 12/Aug/21

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