Question Number 192803 by mokys last updated on 27/May/23
Commented by mokys last updated on 28/May/23
$$?????? \\ $$
Answered by MM42 last updated on 29/May/23
$${Q}\mathrm{1}: \\ $$$$\mathrm{3}+\mathrm{2}{t}+\mathrm{6}{t}−{t}=−\mathrm{4}\Rightarrow{t}=−\mathrm{1}\Rightarrow{p}\left(\mathrm{1},−\mathrm{2},−\mathrm{1}\right) \\ $$$$\overset{\rightarrow} {{u}}_{{L}} =\overset{\rightarrow} {{n}}_{{P}} =\left(\mathrm{1},\mathrm{3},−\mathrm{1}\right)\Rightarrow{L}':\:{x}={t}+\mathrm{1}\:\:\&\:\:{y}=\mathrm{3}{t}−\mathrm{2}\:\:\&\:\:{z}=−{t}−\mathrm{1} \\ $$$${Q}\mathrm{2}: \\ $$$${AB}\left(\mathrm{2},\mathrm{0},−\mathrm{1}\right)\:,\:\:{AC}\left(\mathrm{2},−\mathrm{1},\mathrm{0}\right) \\ $$$$\overset{\rightarrow} {{N}}={A}\overset{\rightarrow} {{B}}×{A}\overset{\rightarrow} {{C}}=\left(\mathrm{1},\mathrm{2},\mathrm{2}\right)\Rightarrow{P}\::\:{x}+\mathrm{2}{y}+\mathrm{2}{z}=\mathrm{0} \\ $$$$\Rightarrow{PH}=\frac{\mid\mathrm{1}+\mathrm{8}\mid}{\:\sqrt{\mathrm{1}+\mathrm{4}+\mathrm{4}}}=\mathrm{1}\:\checkmark \\ $$$${Q}\mathrm{3}: \\ $$$${it}\:{is}\:{not}\:{clear} \\ $$
Commented by MM42 last updated on 29/May/23
Commented by MM42 last updated on 29/May/23
