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Vector-let-L-1-AC-where-A-2-1-3-and-C-1-0-5-and-let-L-2-BD-where-B-1-3-0-and-D-3-4-1-Determine-the-distance-between-L-1-and-L-2-




Question Number 140690 by liberty last updated on 11/May/21
 Vector   let L_1 = AC where A=(2,−1,3) and  C=(1,0,−5)and let L_2 = BD where  B=(1,3,0) and D=(3,−4,1). Determine  the distance between L_1  and L_2 .
$$\:\mathrm{Vector}\: \\ $$$$\mathrm{let}\:\mathcal{L}_{\mathrm{1}} =\:\mathrm{AC}\:\mathrm{where}\:\mathrm{A}=\left(\mathrm{2},−\mathrm{1},\mathrm{3}\right)\:\mathrm{and} \\ $$$$\mathrm{C}=\left(\mathrm{1},\mathrm{0},−\mathrm{5}\right)\mathrm{and}\:\mathrm{let}\:\mathcal{L}_{\mathrm{2}} =\:\mathrm{BD}\:\mathrm{where} \\ $$$$\mathrm{B}=\left(\mathrm{1},\mathrm{3},\mathrm{0}\right)\:\mathrm{and}\:\mathrm{D}=\left(\mathrm{3},−\mathrm{4},\mathrm{1}\right).\:\mathrm{Determine} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathcal{L}_{\mathrm{1}} \:\mathrm{and}\:\mathcal{L}_{\mathrm{2}} . \\ $$
Answered by EDWIN88 last updated on 11/May/21
a vector that is simultaneously perpendicular  to L_1  and L_2  is N =A^→ C ×B^→ D =  determinant (((−1    1    −8)),((   2   −7      1)))  N = −55i −15j +5k   Then d = ∣pr_N  A^→ B ∣ = ∣ A^→ B . (N/(∣N∣)) ∣  d = ((∣(−i+4j−3k).5(−11i−3j+k)∣)/(5(√(121+9+1))))  d = ((∣11−12−3∣)/( (√(131)))) = (4/( (√(131))))
$$\mathrm{a}\:\mathrm{vector}\:\mathrm{that}\:\mathrm{is}\:\mathrm{simultaneously}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathcal{L}_{\mathrm{1}} \:\mathrm{and}\:\mathcal{L}_{\mathrm{2}} \:\mathrm{is}\:\boldsymbol{\mathrm{N}}\:=\overset{\rightarrow} {\mathrm{A}C}\:×\overset{\rightarrow} {\mathrm{B}D}\:=\:\begin{vmatrix}{−\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:−\mathrm{8}}\\{\:\:\:\mathrm{2}\:\:\:−\mathrm{7}\:\:\:\:\:\:\mathrm{1}}\end{vmatrix} \\ $$$$\boldsymbol{\mathrm{N}}\:=\:−\mathrm{55}{i}\:−\mathrm{15}{j}\:+\mathrm{5}{k}\: \\ $$$$\mathrm{Then}\:{d}\:=\:\mid\mathrm{pr}_{\mathrm{N}} \:\overset{\rightarrow} {\mathrm{A}B}\:\mid\:=\:\mid\:\overset{\rightarrow} {\mathrm{A}B}\:.\:\frac{\boldsymbol{\mathrm{N}}}{\mid\boldsymbol{\mathrm{N}}\mid}\:\mid \\ $$$${d}\:=\:\frac{\mid\left(−{i}+\mathrm{4}{j}−\mathrm{3}{k}\right).\mathrm{5}\left(−\mathrm{11}{i}−\mathrm{3}{j}+{k}\right)\mid}{\mathrm{5}\sqrt{\mathrm{121}+\mathrm{9}+\mathrm{1}}} \\ $$$${d}\:=\:\frac{\mid\mathrm{11}−\mathrm{12}−\mathrm{3}\mid}{\:\sqrt{\mathrm{131}}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{131}}} \\ $$

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