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InterviewSolution
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Find the shortest distance between the lines `bar(r) = (4hati - hatj) + lamda(hati + 2hatj - 3hatk)` and ` bar(r) = (hati - hatj + 2hatk) + mu(hati + 4hatj - 5hatk).` |
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Answer» We know that the shortest distance between the lines, `vecr=veca_(1)+lamda vecb_(1)and vecr=veca_(2)+muvecb_(2)` is given as, `=|((veca_(2)-veca_(1))*(vecb_(1)xx vecb_(2)))/(|vecb_(1)xxvecb_(2)|)|` Here, `veca_(1)=4hati-hatj, veca_(2)=hati-hatj+2hatk,` `vecb_(1)=hati+2hatj-3hatk` and `vecb_(2)=hati+4hatj-5hatj` Now, `veca_(2)-veca_(1)=(hati-hatj+2hatk)-(4hati-hatj)` `=-3hati+0hatj+2hatk` `=-3hati+2hatk` and `vecb_(1)xxvecb_(2)=|{:(hati,hatj, hatk),(1,2,-3),(1,4,-5):}|` `=hati(-10+12)-hatj(-5+3)+hatk(4-2)` `=2hati+2hatj+2hatk` `therefore(veca_(2)-veca_(1))*(vecb_(1)xxvecb_(2))=(-3hati+2hatk)*(2hati+2hatj+2hatk)` `=-6+4=-2` and `|vecb_(1)xxvecb_(2)|=sqrt((2)^(2)+(2)^(2)+(2)^(2))=sqrt12` `therefore` Shortest distance, `d=|(-2)/(sqrt12)|` `=(2)/(2sqrt3)=(1)/(sqrt3)` units. |
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