find-the-distance-between-the-planes-2x-y-z-24-and-2x-y-z-36-

Question Number 84218 by Rio Michael last updated on 10/Mar/20

$$\mathrm{find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{planes} \\ $$$$\:\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{y}}−\boldsymbol{{z}}\:=\:\mathrm{24}\:\mathrm{and}\:\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{y}}−\boldsymbol{{z}}\:=\:\mathrm{36} \\ $$

Commented by M±th+et£s last updated on 10/Mar/20

2x−y−z=24⇒⇒2x−y−z−24=0 2x−y−z=36⇒⇒2x−y−z−36=0 h_1 ^_ =h_2 ^_ =( 2^a ,−1^b ,−24^c ) p_1 :(0^x ,0^y ,−24^z ) d=((∣Ax_1 +By_1 +Zz_1 +D∣)/( (√(A^2 +B^2 +C^2 )))) ((∣0+0+24−36∣)/( (√(2^2 +(−1)^2 +(−1)^2 ))))=((12)/( (√6)))=2(√6)

$$\mathrm{2}{x}−{y}−{z}=\mathrm{24}\Rightarrow\Rightarrow\mathrm{2}{x}−{y}−{z}−\mathrm{24}=\mathrm{0} \\ $$$$\mathrm{2}{x}−{y}−{z}=\mathrm{36}\Rightarrow\Rightarrow\mathrm{2}{x}−{y}−{z}−\mathrm{36}=\mathrm{0} \\ $$$$\overset{\_} {{h}}_{\mathrm{1}} =\overset{\_} {{h}}_{\mathrm{2}} =\left(\:\overset{{a}} {\mathrm{2}},−\overset{{b}} {\mathrm{1}},−\mathrm{2}\overset{{c}} {\mathrm{4}}\right) \\ $$$${p}_{\mathrm{1}} :\left(\overset{{x}} {\mathrm{0}},\overset{{y}} {\mathrm{0}},−\mathrm{2}\overset{{z}} {\mathrm{4}}\right) \\ $$$${d}=\frac{\mid{Ax}_{\mathrm{1}} +{By}_{\mathrm{1}} +{Zz}_{\mathrm{1}} +{D}\mid}{\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} }} \\ $$$$\frac{\mid\mathrm{0}+\mathrm{0}+\mathrm{24}−\mathrm{36}\mid}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{6}}}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$

Answered by mr W last updated on 10/Mar/20

distance from origin to plane 1: d_1 =((24)/( (√(2^2 +1^2 +1^2 ))))=((24)/( (√6))) distance from origin to plane 2: d_2 =((36)/( (√(2^2 +1^2 +1^2 ))))=((36)/( (√6))) distance between the planes: d_2 −d_1 =((36−24)/( (√6)))=2(√6)

$${distance}\:{from}\:{origin}\:{to}\:{plane}\:\mathrm{1}: \\ $$$${d}_{\mathrm{1}} =\frac{\mathrm{24}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{24}}{\:\sqrt{\mathrm{6}}} \\ $$$${distance}\:{from}\:{origin}\:{to}\:{plane}\:\mathrm{2}: \\ $$$${d}_{\mathrm{2}} =\frac{\mathrm{36}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{36}}{\:\sqrt{\mathrm{6}}} \\ $$$${distance}\:{between}\:{the}\:{planes}: \\ $$$${d}_{\mathrm{2}} −{d}_{\mathrm{1}} =\frac{\mathrm{36}−\mathrm{24}}{\:\sqrt{\mathrm{6}}}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$

Answered by mr W last updated on 10/Mar/20

find any point on plane 1, e.g. x=12,y=0,z=0 distance from this point to plane 2 is the distance between the planes: ((∣2×12−36∣)/( (√(2^2 +1^2 +1^2 ))))=((12)/( (√6)))=2(√6)

$${find}\:{any}\:{point}\:{on}\:{plane}\:\mathrm{1},\:{e}.{g}. \\ $$$${x}=\mathrm{12},{y}=\mathrm{0},{z}=\mathrm{0} \\ $$$${distance}\:{from}\:{this}\:{point}\:{to}\:{plane}\:\mathrm{2} \\ $$$${is}\:{the}\:{distance}\:{between}\:{the}\:{planes}: \\ $$$$\frac{\mid\mathrm{2}×\mathrm{12}−\mathrm{36}\mid}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{6}}}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$

Commented by Rio Michael last updated on 10/Mar/20