Find the equation of the plane thrputh the line x-1/3= y-4/2 = z-4/-2 and parallel to the line x+1/2 = 1-y/4 = z+2/1.  Hence find the shortest distance between the lines.

Asked by ponnarasan98 | 15th Mar, 2019, 07:27: PM

Expert Answer:

Equation of  line passing through the plane  is  begin mathsize 12px style fraction numerator x minus 1 over denominator 3 end fraction space equals space fraction numerator y minus 4 over denominator 2 end fraction equals fraction numerator z minus 4 over denominator negative 2 end fraction end style  ..........................(1)
since the line passes through the point (1,4,4), this point lies in the plane.
 
Hence equation of the plane is  a(x-1)+b(y-4)+c(z-4) = 0 ................................(2)
 
where a,b and c are direction ratios of the normal to the plane.
 
The plane is parallel to the line begin mathsize 12px style fraction numerator x plus 1 over denominator 2 end fraction space equals space fraction numerator y minus 1 over denominator negative 4 end fraction equals fraction numerator z plus 2 over denominator 1 end fraction end style .........................(3)
Hence normal to the plane is perpendicular to dircetion of both the vectors associated to line (1) and (3).
 
Hence normal vector of the plane is obtained from cross product of vectors associated to line (1) and (3)
 
normal vector = begin mathsize 12px style open vertical bar table row cell i with hat on top end cell cell j with hat on top end cell cell k with hat on top end cell row 3 2 cell negative 2 end cell row 2 cell negative 4 end cell 1 end table close vertical bar space equals space minus 6 i space minus 7 j space minus 16 k end style
hence the direction ratios of normal are 6, 7, 16 . we get the equation of plane by substituting these direction ratios in eqn.(2)
 
a(x-1)+b(y-4)+c(z-4) = 6(x-1)+7(y-4)+16(z-4) = 6x+7y+16z-98 = 0 ............................(4)
 
The line given by eqn.(3) passes through the point (-1, 1, -2).
Hence distance between the lines is same as the perpendicular distance from this point to the plane given by eqn.(4)
 
Perpendicular distance from a point (x1 , y1 , z1 ) to the plane ax+by+cz-d=0 is begin mathsize 12px style open vertical bar fraction numerator a space x subscript 1 space plus space b space y subscript 1 space plus space c space z subscript 1 space minus d over denominator square root of a squared plus b squared plus c squared end root end fraction close vertical bar space end style
hence perpendicular distance = begin mathsize 12px style open vertical bar fraction numerator 6 open parentheses negative 1 close parentheses space plus 7 open parentheses 1 close parentheses space plus 16 open parentheses negative 2 close parentheses space minus 98 over denominator square root of 36 plus 49 plus 256 end root end fraction close vertical bar space equals space fraction numerator 129 over denominator square root of 341 end fraction end style

Answered by Thiyagarajan K | 16th Mar, 2019, 05:30: PM

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