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!

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

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