# SOLUTION: Find the shortest distance between the plane 2x+4y-2z=6 and the line (x,y,z) = (-1,-2,4) + t(-2,-1,-4). Distance =???? Thank you!

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: Find the shortest distance between the plane 2x+4y-2z=6 and the line (x,y,z) = (-1,-2,4) + t(-2,-1,-4). Distance =???? Thank you!      Log On

 Algebra: Linear Algebra (NOT Linear Equations) Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Linear Algebra Question 62447: Find the shortest distance between the plane 2x+4y-2z=6 and the line (x,y,z) = (-1,-2,4) + t(-2,-1,-4). Distance =???? Thank you!Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!Find the shortest distance between the plane(P SAY) 2x+4y-2z=6 and the line(L SAY) (x,y,z) = (Q SAY)(-1,-2,4) + t(-2,-1,-4). Distance =???? IF L IS NOT PARALLEL TO P THEN S.D=0 AS IT INTERSECTS THE PLANE. IF L||P NORMAL TO PLANE IS PERPENDICULAR TO LINE DRS OF NORMAL TO PLANE ...(2,4,-2) DRS OF LINE ....(-2,-1,-4) 2*-2+4*-1+-2*-4=-4-4+8=0...HENCE L||P SO TAKE ANY POINT ON LINE AND FIND ITS PERPENDICULAR DISTANCE FROM THE PLANE. ANY POINT ON LINE IS (-1,-2,-4) S.D = [2*-1+4*-2+(-2)(-4)-6]/SQRT(2^2+4^2+2^2) = -8/SQRT(24)=-2SQRT(6)/3