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Find-the-distance-from-the-point-S-1-1-5-to-the-line-L-x-1-t-y-3-t-z-2t-

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Question Number 123876 by john_santu last updated on 29/Nov/20
Find the distance from the point S(1,1,5) to the line L : { ((x=1+t)),((y=3−t )),((z=2t)) :}.
FindthedistancefromthepointS(1,1,5)tothelineL:{x=1+ty=3−tz=2t.
Answered by mr W last updated on 29/Nov/20
Method I d^2 =(1+t−1)^2 +(3−t−1)^2 +(2t−5)^2 =6t^2 −24t+29 ((d(d^2 ))/dt)=12t−24=0 ⇒t=2 d_(min) ^2 =6×2^2 −24×2+29=5 ⇒distance=d_(min) =(√5)
MethodId2=(1+t−1)2+(3−t−1)2+(2t−5)2=6t2−24t+29d(d2)dt=12t−24=0⇒t=2dmin2=6×22−24×2+29=5⇒distance=dmin=5
Answered by mr W last updated on 29/Nov/20
Method II S(1,1,5) P(1,3,0) PQ=(1,−1,2) PS=(0,2,−5) distance=(√(∣PS∣^2 −(((PQ•PS)/(∣PQ∣)))^2 )) =(√(0^2 +2^2 +(−5)^2 −(((1×0−1×2−2×5)^2 )/(1^2 +(−1)^2 +2^2 )))) =(√(29−24)) =(√5)
MethodIIS(1,1,5)P(1,3,0)PQ=(1,−1,2)PS=(0,2,−5)distance=∣PS∣2−(PQ∙PS∣PQ∣)2=02+22+(−5)2−(1×0−1×2−2×5)212+(−1)2+22=29−24=5
Answered by mr W last updated on 29/Nov/20
Method III S(1,1,5) P(1,3,0) SP=(0,2,−5) PQ=(1,−1,2) distance=((∣SP×PQ∣)/(∣PQ∣))=((∣(0,2,−5)×(1,−1,2)∣)/( (√(1^2 +(−1)^2 +2^2 )))) =((√((−1)^2 +(−5)^2 +(−2)^2 ))/( (√6))) =(√((30)/6)) =(√5)
MethodIIIS(1,1,5)P(1,3,0)SP=(0,2,−5)PQ=(1,−1,2)distance=∣SP×PQ∣∣PQ∣=∣(0,2,−5)×(1,−1,2)∣12+(−1)2+22=(−1)2+(−5)2+(−2)26=306=5

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