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Question Number 96240 by mr W last updated on 30/May/20
Find the shortest distance from the point P(2,−3,5) to the line L ((x+3)/2)=((y−1)/(−3))=((z−2)/4)
$${Find}\:{the}\:{shortest}\:{distance}\:{from}\:{the} \\ $$$${point}\:{P}\left(\mathrm{2},−\mathrm{3},\mathrm{5}\right)\:{to}\:{the}\:{line}\:{L} \\ $$$$\frac{{x}+\mathrm{3}}{\mathrm{2}}=\frac{{y}−\mathrm{1}}{−\mathrm{3}}=\frac{{z}−\mathrm{2}}{\mathrm{4}} \\ $$
Answered by 1549442205 last updated on 31/May/20
Denote Q be the point on L such that PQ⊥L⇒Q(2t−3;−3t+1;4t+2).Then PQ^(→) =(2t−5;−3t+4;4t−3).The direction vector of L is p^(→) =(2;−3;4) perpemdicular to PQ^(→) .Hence,p^(→) .PQ^(→) =0⇔2(2t−5)+3(3t−4) +4(4t−3)⇔29t−34=0⇒t=((34)/(29)).From that PQ^(→) =(((−77)/(29));((14)/(29));((49)/(29)))⇒PQ=((7(√(11^2 +2^2 +7^2 )))/(29))= ((7(√(174)))/(29)) is shortest distance from P to L
$$\mathrm{Denote}\:\mathrm{Q}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{L}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{PQ}\bot\mathrm{L}\Rightarrow\mathrm{Q}\left(\mathrm{2t}−\mathrm{3};−\mathrm{3t}+\mathrm{1};\mathrm{4t}+\mathrm{2}\right).\mathrm{Then} \\ $$$$\overset{\rightarrow} {\mathrm{PQ}}=\left(\mathrm{2t}−\mathrm{5};−\mathrm{3t}+\mathrm{4};\mathrm{4t}−\mathrm{3}\right).\mathrm{The}\:\mathrm{direction}\: \\ $$$$\mathrm{vector}\:\mathrm{of}\:\:\mathrm{L}\:\mathrm{is}\:\overset{\rightarrow} {\mathrm{p}}=\left(\mathrm{2};−\mathrm{3};\mathrm{4}\right)\:\mathrm{perpemdicular} \\ $$$$\mathrm{to}\:\overset{\rightarrow} {\mathrm{PQ}}.\mathrm{Hence},\overset{\rightarrow} {\mathrm{p}}.\overset{\rightarrow} {\mathrm{PQ}}=\mathrm{0}\Leftrightarrow\mathrm{2}\left(\mathrm{2t}−\mathrm{5}\right)+\mathrm{3}\left(\mathrm{3t}−\mathrm{4}\right) \\ $$$$+\mathrm{4}\left(\mathrm{4t}−\mathrm{3}\right)\Leftrightarrow\mathrm{29t}−\mathrm{34}=\mathrm{0}\Rightarrow\mathrm{t}=\frac{\mathrm{34}}{\mathrm{29}}.\mathrm{From}\:\mathrm{that} \\ $$$$\overset{\rightarrow} {\mathrm{PQ}}=\left(\frac{−\mathrm{77}}{\mathrm{29}};\frac{\mathrm{14}}{\mathrm{29}};\frac{\mathrm{49}}{\mathrm{29}}\right)\Rightarrow\mathrm{PQ}=\frac{\mathrm{7}\sqrt{\mathrm{1}\overset{\mathrm{2}} {\mathrm{1}}+\mathrm{2}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }}{\mathrm{29}}= \\ $$$$\frac{\mathrm{7}\sqrt{\mathrm{174}}}{\mathrm{29}}\:\mathrm{is}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{L} \\ $$
Commented by mr W last updated on 31/May/20
thank you sir!
$${thank}\:{you}\:{sir}! \\ $$
Answered by john santu last updated on 31/May/20
let S(2s−3, −3s+1, 5s+2) a point on line L . take vector PS = (2s−5,−3s+4,5s−3) PS ⊥n=(2,−3,4) 4s−10+9s−12+20s−12=0 33s = 34 ⇒s = ((34)/(33)) PS(−((97)/(33)) , ((30)/(33)), ((71)/(33))). ∣PS∣ = ((√(15,350))/(33)) ≈ 3.754397483
$$\mathrm{let}\:\mathrm{S}\left(\mathrm{2s}−\mathrm{3},\:−\mathrm{3s}+\mathrm{1},\:\mathrm{5s}+\mathrm{2}\right)\:\mathrm{a}\:\mathrm{point}\: \\ $$$$\mathrm{on}\:\mathrm{line}\:\mathrm{L}\:.\:\mathrm{take}\:\mathrm{vector}\:\mathrm{PS}\:=\:\left(\mathrm{2s}−\mathrm{5},−\mathrm{3s}+\mathrm{4},\mathrm{5s}−\mathrm{3}\right) \\ $$$$\boldsymbol{\mathrm{PS}}\:\bot\mathrm{n}=\left(\mathrm{2},−\mathrm{3},\mathrm{4}\right)\: \\ $$$$\mathrm{4s}−\mathrm{10}+\mathrm{9s}−\mathrm{12}+\mathrm{20s}−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{33s}\:=\:\mathrm{34}\:\Rightarrow\mathrm{s}\:=\:\frac{\mathrm{34}}{\mathrm{33}} \\ $$$$\mathrm{PS}\left(−\frac{\mathrm{97}}{\mathrm{33}}\:,\:\frac{\mathrm{30}}{\mathrm{33}},\:\frac{\mathrm{71}}{\mathrm{33}}\right). \\ $$$$\mid\mathrm{PS}\mid\:=\:\frac{\sqrt{\mathrm{15},\mathrm{350}}}{\mathrm{33}}\:\approx\:\mathrm{3}.\mathrm{754397483} \\ $$$$ \\ $$
Commented by mr W last updated on 31/May/20
thank you sir! typo in first line 5s+2 ? s=((34)/(29)) ?
$${thank}\:{you}\:{sir}! \\ $$$${typo}\:{in}\:{first}\:{line}\:\mathrm{5}{s}+\mathrm{2}\:? \\ $$$${s}=\frac{\mathrm{34}}{\mathrm{29}}\:? \\ $$
Commented by bobhans last updated on 31/May/20
mr santu typo in 5s+2 it should be 4s+2
$$\mathrm{mr}\:\mathrm{santu}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{5s}+\mathrm{2}\: \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{4s}+\mathrm{2} \\ $$
Answered by mr W last updated on 31/May/20
METHOD I any point on line Q: x=−3+2k y=1−3k z=2+4k distance PQ=d D=d^2 =(2+3−2k)^2 +(−3−1+3k)^2 +(5−2−4k)^2 D=(2k−5)^2 +(3k−4)^2 +(4k−3)^2 such that d is minimum: (dD/dk)=2(2k−5)2+2(3k−4)3+2(4k−3)4=0 ⇒29k=34 ⇒k=((34)/(29)) d_(min) =(√((2×((34)/(29))−5)^2 +(3×((34)/(29))−4)^2 +(4×((34)/(29))−3)^2 ))=((7(√(174)))/(29)) METHOD II known point on line R(−3,1,2) PR^(→) =(5,−4,3) l^→ =(2,−3,4) cos θ=((PR∙l)/(∣PR∣∣l∣)) d=∣PR∣ sin θ=∣PR∣(√(1−(((PR∙l)/(∣PR∣∣l∣)))^2 )) d=(√(∣PR∣^2 −(((PR∙l)^2 )/(∣l∣^2 )))) d=(√(5^2 +4^2 +3^2 −(((5×2+4×3+3×4)^2 )/(2^2 +3^2 +4^2 )))) d=(√(50−((34^2 )/(29))))=((7(√(174)))/(29)) METHOD III plane perpendicular to line L is 2x−3y+4z+s=0 point P is on the plane: 2×2−3×(−3)+4×5+s=0 s=−33 ⇒2x−3y+4z−33=0 intersection of line with plane at Q: x=−3+2k y=1−3k z=2+4k 2(−3+2k)−3(1−3k)+4(2+4k)−33=0 29k−34=0 ⇒k=((34)/(29)) PQ=(√((2+3−2k)^2 +(−3−1+3k)^2 +(5−2−4k)^2 )) PQ=(√((5−2k)^2 +(−4+3k)^2 +(3−4k)^2 )) PQ=(√((5−2×((34)/(29)))^2 +(−4+3×((34)/(29)))^2 +(3−4×((34)/(29)))^2 )) d_(min) =PQ=((7(√(174)))/(29)) METHOD IV point on line Q(−3+2k, 1−3k, 2+4k) l^→ =(2,−3,4) PQ^(→) =(2+3−2k,−3−1+3k,5−2−4k) PQ^(→) =(5−2k,−4+3k,3−4k) such that PQ^(→) ⊥l^(→) : (5−2k)×2×(−4+3k)×(−3)+(3−4k)×4=0 ⇒k=((34)/(29)) d_(min) =∣PQ∣=(√((5−2×((34)/(29)))^2 +(−4+3×((34)/(29)))^2 +(3−4×((34)/(29)))^2 )) =((7(√(174)))/(29)) METHOD V known point on line R(−3,1,2) PR^(→) =(5,−4,3) l^→ =(2,−3,4) ∣l^→ ∣=(√(2^2 +3^2 +4^2 ))=(√(29)) PR^(→) ×l^→ =(5,−4,3)×(2,−3,4)=(−7,−14,−7) ∣PR^(→) ×l^→ ∣=(√(7^2 +14^2 +7^2 ))=7(√6) d=((∣PR×l^→ ∣)/(∣l^→ ∣))=((7(√6))/( (√(29))))=((7(√(174)))/(29))
$$\boldsymbol{{METHOD}}\:\:\boldsymbol{{I}} \\ $$$${any}\:{point}\:{on}\:{line}\:{Q}: \\ $$$${x}=−\mathrm{3}+\mathrm{2}{k} \\ $$$${y}=\mathrm{1}−\mathrm{3}{k} \\ $$$${z}=\mathrm{2}+\mathrm{4}{k} \\ $$$${distance}\:{PQ}={d} \\ $$$${D}={d}^{\mathrm{2}} =\left(\mathrm{2}+\mathrm{3}−\mathrm{2}{k}\right)^{\mathrm{2}} +\left(−\mathrm{3}−\mathrm{1}+\mathrm{3}{k}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{2}−\mathrm{4}{k}\right)^{\mathrm{2}} \\ $$$${D}=\left(\mathrm{2}{k}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{3}{k}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{4}{k}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$${such}\:{that}\:{d}\:{is}\:{minimum}: \\ $$$$\frac{{dD}}{{dk}}=\mathrm{2}\left(\mathrm{2}{k}−\mathrm{5}\right)\mathrm{2}+\mathrm{2}\left(\mathrm{3}{k}−\mathrm{4}\right)\mathrm{3}+\mathrm{2}\left(\mathrm{4}{k}−\mathrm{3}\right)\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{29}{k}=\mathrm{34} \\ $$$$\Rightarrow{k}=\frac{\mathrm{34}}{\mathrm{29}} \\ $$$${d}_{{min}} =\sqrt{\left(\mathrm{2}×\frac{\mathrm{34}}{\mathrm{29}}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{3}×\frac{\mathrm{34}}{\mathrm{29}}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{4}×\frac{\mathrm{34}}{\mathrm{29}}−\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{7}\sqrt{\mathrm{174}}}{\mathrm{29}} \\ $$$$ \\ $$$$\boldsymbol{{METHOD}}\:\:\boldsymbol{{II}} \\ $$$${known}\:{point}\:{on}\:{line}\:\boldsymbol{{R}}\left(−\mathrm{3},\mathrm{1},\mathrm{2}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{PR}}}=\left(\mathrm{5},−\mathrm{4},\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{l}}}=\left(\mathrm{2},−\mathrm{3},\mathrm{4}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\boldsymbol{{PR}}\centerdot\boldsymbol{{l}}}{\mid\boldsymbol{{PR}}\mid\mid\boldsymbol{{l}}\mid} \\ $$$${d}=\mid\boldsymbol{{PR}}\mid\:\mathrm{sin}\:\theta=\mid\boldsymbol{{PR}}\mid\sqrt{\mathrm{1}−\left(\frac{\boldsymbol{{PR}}\centerdot\boldsymbol{{l}}}{\mid\boldsymbol{{PR}}\mid\mid\boldsymbol{{l}}\mid}\right)^{\mathrm{2}} } \\ $$$${d}=\sqrt{\mid\boldsymbol{{PR}}\mid^{\mathrm{2}} −\frac{\left(\boldsymbol{{PR}}\centerdot\boldsymbol{{l}}\right)^{\mathrm{2}} }{\mid\boldsymbol{{l}}\mid^{\mathrm{2}} }} \\ $$$${d}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\frac{\left(\mathrm{5}×\mathrm{2}+\mathrm{4}×\mathrm{3}+\mathrm{3}×\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }} \\ $$$${d}=\sqrt{\mathrm{50}−\frac{\mathrm{34}^{\mathrm{2}} }{\mathrm{29}}}=\frac{\mathrm{7}\sqrt{\mathrm{174}}}{\mathrm{29}} \\ $$$$ \\ $$$$\boldsymbol{{METHOD}}\:\:\boldsymbol{{III}} \\ $$$${plane}\:{perpendicular}\:{to}\:{line}\:{L}\:{is} \\ $$$$\mathrm{2}{x}−\mathrm{3}{y}+\mathrm{4}{z}+{s}=\mathrm{0} \\ $$$$ \\ $$$${point}\:{P}\:{is}\:{on}\:{the}\:{plane}: \\ $$$$\mathrm{2}×\mathrm{2}−\mathrm{3}×\left(−\mathrm{3}\right)+\mathrm{4}×\mathrm{5}+{s}=\mathrm{0} \\ $$$${s}=−\mathrm{33} \\ $$$$\Rightarrow\mathrm{2}{x}−\mathrm{3}{y}+\mathrm{4}{z}−\mathrm{33}=\mathrm{0} \\ $$$$ \\ $$$${intersection}\:{of}\:{line}\:{with}\:{plane}\:{at}\:{Q}: \\ $$$${x}=−\mathrm{3}+\mathrm{2}{k} \\ $$$${y}=\mathrm{1}−\mathrm{3}{k} \\ $$$${z}=\mathrm{2}+\mathrm{4}{k} \\ $$$$\mathrm{2}\left(−\mathrm{3}+\mathrm{2}{k}\right)−\mathrm{3}\left(\mathrm{1}−\mathrm{3}{k}\right)+\mathrm{4}\left(\mathrm{2}+\mathrm{4}{k}\right)−\mathrm{33}=\mathrm{0} \\ $$$$\mathrm{29}{k}−\mathrm{34}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{34}}{\mathrm{29}} \\ $$$${PQ}=\sqrt{\left(\mathrm{2}+\mathrm{3}−\mathrm{2}{k}\right)^{\mathrm{2}} +\left(−\mathrm{3}−\mathrm{1}+\mathrm{3}{k}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{2}−\mathrm{4}{k}\right)^{\mathrm{2}} } \\ $$$${PQ}=\sqrt{\left(\mathrm{5}−\mathrm{2}{k}\right)^{\mathrm{2}} +\left(−\mathrm{4}+\mathrm{3}{k}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{4}{k}\right)^{\mathrm{2}} } \\ $$$${PQ}=\sqrt{\left(\mathrm{5}−\mathrm{2}×\frac{\mathrm{34}}{\mathrm{29}}\right)^{\mathrm{2}} +\left(−\mathrm{4}+\mathrm{3}×\frac{\mathrm{34}}{\mathrm{29}}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{4}×\frac{\mathrm{34}}{\mathrm{29}}\right)^{\mathrm{2}} } \\ $$$${d}_{{min}} ={PQ}=\frac{\mathrm{7}\sqrt{\mathrm{174}}}{\mathrm{29}} \\ $$$$ \\ $$$$\boldsymbol{{METHOD}}\:\:\boldsymbol{{IV}} \\ $$$${point}\:{on}\:{line}\:{Q}\left(−\mathrm{3}+\mathrm{2}{k},\:\mathrm{1}−\mathrm{3}{k},\:\mathrm{2}+\mathrm{4}{k}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{l}}}=\left(\mathrm{2},−\mathrm{3},\mathrm{4}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{PQ}}}=\left(\mathrm{2}+\mathrm{3}−\mathrm{2}{k},−\mathrm{3}−\mathrm{1}+\mathrm{3}{k},\mathrm{5}−\mathrm{2}−\mathrm{4}{k}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{PQ}}}=\left(\mathrm{5}−\mathrm{2}{k},−\mathrm{4}+\mathrm{3}{k},\mathrm{3}−\mathrm{4}{k}\right) \\ $$$${such}\:{that}\:\overset{\rightarrow} {\boldsymbol{{PQ}}}\bot\overset{\rightarrow} {\boldsymbol{{l}}}: \\ $$$$\left(\mathrm{5}−\mathrm{2}{k}\right)×\mathrm{2}×\left(−\mathrm{4}+\mathrm{3}{k}\right)×\left(−\mathrm{3}\right)+\left(\mathrm{3}−\mathrm{4}{k}\right)×\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{34}}{\mathrm{29}} \\ $$$${d}_{{min}} =\mid\boldsymbol{{PQ}}\mid=\sqrt{\left(\mathrm{5}−\mathrm{2}×\frac{\mathrm{34}}{\mathrm{29}}\right)^{\mathrm{2}} +\left(−\mathrm{4}+\mathrm{3}×\frac{\mathrm{34}}{\mathrm{29}}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{4}×\frac{\mathrm{34}}{\mathrm{29}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{7}\sqrt{\mathrm{174}}}{\mathrm{29}} \\ $$$$ \\ $$$$\boldsymbol{{METHOD}}\:\:\boldsymbol{{V}} \\ $$$${known}\:{point}\:{on}\:{line}\:\boldsymbol{{R}}\left(−\mathrm{3},\mathrm{1},\mathrm{2}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{PR}}}=\left(\mathrm{5},−\mathrm{4},\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{l}}}=\left(\mathrm{2},−\mathrm{3},\mathrm{4}\right) \\ $$$$\mid\overset{\rightarrow} {\boldsymbol{{l}}}\mid=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{29}} \\ $$$$\overset{\rightarrow} {\boldsymbol{{PR}}}×\overset{\rightarrow} {\boldsymbol{{l}}}=\left(\mathrm{5},−\mathrm{4},\mathrm{3}\right)×\left(\mathrm{2},−\mathrm{3},\mathrm{4}\right)=\left(−\mathrm{7},−\mathrm{14},−\mathrm{7}\right) \\ $$$$\mid\overset{\rightarrow} {\boldsymbol{{PR}}}×\overset{\rightarrow} {\boldsymbol{{l}}}\mid=\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{14}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\mathrm{7}\sqrt{\mathrm{6}} \\ $$$${d}=\frac{\mid\boldsymbol{{PR}}×\overset{\rightarrow} {\boldsymbol{{l}}}\mid}{\mid\overset{\rightarrow} {\boldsymbol{{l}}}\mid}=\frac{\mathrm{7}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{29}}}=\frac{\mathrm{7}\sqrt{\mathrm{174}}}{\mathrm{29}} \\ $$
Commented by bobhans last updated on 31/May/20
waw====great
$$\mathrm{waw}====\mathrm{great} \\ $$
Commented by mr W last updated on 31/May/20
Method II is my favourite method d_(min) =(√(∣RP^(→) ∣^2 −(((RP^(→) •l^(→) )^2 )/(∣l^(→) ∣^2 )))) this formula is not teached in school and in book.
$${Method}\:{II}\:{is}\:{my}\:{favourite}\:{method} \\ $$$${d}_{{min}} =\sqrt{\mid\overset{\rightarrow} {\boldsymbol{{RP}}}\mid^{\mathrm{2}} −\frac{\left(\overset{\rightarrow} {\boldsymbol{{RP}}}\:\bullet\overset{\rightarrow} {\boldsymbol{{l}}}\right)^{\mathrm{2}} }{\mid\overset{\rightarrow} {\boldsymbol{{l}}}\mid^{\mathrm{2}} }} \\ $$$${this}\:{formula}\:{is}\:{not}\:{teached}\:{in}\:{school} \\ $$$${and}\:{in}\:{book}. \\ $$
Commented by mr W last updated on 31/May/20
Commented by bobhans last updated on 31/May/20
whether the point R is arbitrarily provided it lies on the L line
Commented by bobhans last updated on 31/May/20
this method from Pytagorean theorem?
$$\mathrm{this}\:\mathrm{method}\:\mathrm{from}\:\mathrm{Pytagorean}\:\mathrm{theorem}? \\ $$

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