Mensuration Questions

Question Number 105404 by bemath last updated on 28/Jul/20

$${Cube}\:{ABCD}.{EFGH}\:{with} \\$$$${length}\:{side}\:\mathrm{2}\:{cm}.\:{Point}\:{P}\:{is}\:{the} \\$$$${center}\:{of}\:{ABFE}\:{plane}.\:{The} \\$$$${distance}\:{of}\:{HP}\:{line}\:{and}\:{the}\:{BG} \\$$$${line}\:{is}\:\_\_\: \\$$

Answered by john santu last updated on 28/Jul/20

Commented by john santu last updated on 28/Jul/20

$${normal}\:{vector}\:{of}\:{plane}\:{BGP}\:' \\$$$${B}\left(\mathrm{2},\mathrm{2},\mathrm{0}\right);\:{P}\:'\left(\mathrm{2},\mathrm{3},\mathrm{1}\right);\:{G}\left(\mathrm{0},\mathrm{2},\mathrm{2}\right) \\$$$${with}\:{BG}=\left(−\mathrm{2},\mathrm{0},\mathrm{2}\right),\:{BP}\:'=\left(\mathrm{0},\mathrm{1},\mathrm{1}\right) \\$$$$\overset{\rightarrow} {{n}}=\begin{vmatrix}{−\mathrm{2}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix}=\:\left(−\mathrm{2},\mathrm{2},−\mathrm{2}\right) \\$$$${equation}\:{of}\:{plane}\:{BP}\:'{G}\: \\$$$$−\mathrm{2}{x}+\mathrm{2}{y}−\mathrm{2}{z}\:\:=\:\mathrm{0}\:; \\$$$${put}\:{point}\:{H}\left(\mathrm{0},\mathrm{0},\mathrm{2}\right)\: \\$$$${distance}\:{line}\:{of}\:{HP}\:{to}\:{line}\:{of} \\$$$${BG}\:=\:\mid\frac{−\mathrm{4}}{\sqrt{\mathrm{4}+\mathrm{4}+\mathrm{4}}}\mid\:=\:\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}} \\$$

Commented by john santu last updated on 28/Jul/20

$${if}\:{put}\:{point}\:{P}\left(\mathrm{2},\mathrm{1},\mathrm{1}\right) \\$$$${distance}\:=\:\mid\frac{−\mathrm{4}+\mathrm{2}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\mid\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}} \\$$