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Find the point on the plane 4x+3y+z=2 that is closest to (1,-1,1).
Dear Akif Indirectly we hwve to find out foot of perpendicular from (1, - 1,1) to the given plane x -1 /4 = y +1/3 = z – 1/1 = – 2(0) / 26 x -1 /4 = y +1/3 = z – 1/1 = 0 hence x =1, y = -1, z = 1 Means (1, – 1, 1) lies on givne plane
Dear student The normal vector to the plane is (4,3,1)P = (1,-1,1) + c(4,3,1) = (1+4c, -1+3c , 1+c) 4(1+4c) +3(-1+3c) + 1+c = 126c +2 = 1 c = -1/26 P = (1,-1,1) - 1/26 ( 4,3,1) P = (0.84 , -1.11 , 0.96)
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