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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 | 15 Mar, 2019, 07:27: PM
Equation of  line passing through the plane  is    ..........................(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 .........................(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 =
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
hence perpendicular distance =
Answered by Thiyagarajan K | 16 Mar, 2019, 05:30: PM

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