Find the point where the line passing through 1 0 0 and perp

Find the point where the line passing through (1, 0, 0) and perpendicular to the

plane 2x + 3y + z = 0 meets the plane 2x + 3y + z = 12.

Solution

LINE THROUGH [ 1,0,0] IS GIVEN BY [(X-1) / A ] = [ Y / B ] = [ Z / C ] = T……………………..1 PLANE IS …. 2X+3Y+Z = 0……………………………………………2 DRS OF NORMAL TO PLANE ARE …..[ 2 , 3 , 1 ] LINE IS PERPENDICULAR TO THE PLANE … THAT IS LINE IS PARALLEL TO THE NORMAL TO THE PLANE SO DRS OF LINE ARE ….. [ 2 , 3 , 1 ] = [ A , B , C ] HENCE EQN. OF LINE IS ….. [(X-1) / 2 ] = [ Y / 3 ] = [ Z / 1 ] = T……………………..2 X-1=2T…….X= 1+ 2T Y=3T Z=T IT MEETS THE PLANE …. 2X+3Y+Z = 12 AT 2[1+2T]+3[3T]+T = 12 T = 5 / 7 HENCE THE MEETING POINT OF LINE WITH PLANE IS …… X= 1+ 2T     = 2*(5/7)+1     = 2.4286 Y=3T     = 3*5 / 7     = 2.1429 Z=T     = 5 / 7     = 0.7143