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Question-53025




Question Number 53025 by ajfour last updated on 16/Jan/19
Commented by ajfour last updated on 16/Jan/19
Find minimum distance of line  AB from cylinder axis.  (correction: the shown segment  with which AB makes ∠ α should   end in foot of vertical from A.)
$${Find}\:{minimum}\:{distance}\:{of}\:{line} \\ $$$${AB}\:{from}\:{cylinder}\:{axis}. \\ $$$$\left({correction}:\:{the}\:{shown}\:{segment}\right. \\ $$$${with}\:{which}\:{AB}\:{makes}\:\angle\:\alpha\:{should} \\ $$$$\left.\:{end}\:{in}\:{foot}\:{of}\:{vertical}\:{from}\:{A}.\right) \\ $$$$ \\ $$
Answered by mr W last updated on 16/Jan/19
min. d_⊥ =(√(R^2 −((1/2)×((2R)/(tan α)))^2 ))  =R(√(1−(1/(tan^2  α))))
$${min}.\:{d}_{\bot} =\sqrt{{R}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{R}}{\mathrm{tan}\:\alpha}\right)^{\mathrm{2}} } \\ $$$$={R}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}} \\ $$
Commented by ajfour last updated on 16/Jan/19
Very imaginative Sir! Thanks, i  had solved it the tedious vector way.
$${Very}\:{imaginative}\:{Sir}!\:{Thanks},\:{i} \\ $$$${had}\:{solved}\:{it}\:{the}\:{tedious}\:{vector}\:{way}. \\ $$
Commented by mr W last updated on 16/Jan/19
construct a cylinder with the same  axis, if this cylinder touches the line  AB at only one point, then the radius of  this cylinder is the minimum distance  searched.  it′s obvious that this cylinder  touches the line AB at its midpoint.  AB′=(h/(tan α))=((2R)/(tan α))  AM′=((AB′)/2)=(R/(tan α))  d_⊥ =OM′=(√(R^2 −((R/(tan α)))^2 ))=R(√(1−(1/(tan^2  α))))
$${construct}\:{a}\:{cylinder}\:{with}\:{the}\:{same} \\ $$$${axis},\:{if}\:{this}\:{cylinder}\:{touches}\:{the}\:{line} \\ $$$${AB}\:{at}\:{only}\:{one}\:{point},\:{then}\:{the}\:{radius}\:{of} \\ $$$${this}\:{cylinder}\:{is}\:{the}\:{minimum}\:{distance} \\ $$$${searched}.\:\:{it}'{s}\:{obvious}\:{that}\:{this}\:{cylinder} \\ $$$${touches}\:{the}\:{line}\:{AB}\:{at}\:{its}\:{midpoint}. \\ $$$${AB}'=\frac{{h}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{2}{R}}{\mathrm{tan}\:\alpha} \\ $$$${AM}'=\frac{{AB}'}{\mathrm{2}}=\frac{{R}}{\mathrm{tan}\:\alpha} \\ $$$${d}_{\bot} ={OM}'=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{tan}\:\alpha}\right)^{\mathrm{2}} }={R}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}} \\ $$
Commented by ajfour last updated on 16/Jan/19
correct Sir, dint follow..
$${correct}\:{Sir},\:{dint}\:{follow}.. \\ $$
Commented by mr W last updated on 16/Jan/19
Commented by Otchere Abdullai last updated on 16/Jan/19
wow! this man  have creative mind his   solutions are always correct and  unique
$${wow}!\:{this}\:{man}\:\:{have}\:{creative}\:{mind}\:{his}\: \\ $$$${solutions}\:{are}\:{always}\:{correct}\:{and} \\ $$$${unique} \\ $$

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