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Find the Point on the plane 4x+3y+z=2 that is Closest to (1,-1,1).

Find the Point on the plane 4x+3y+z=2 that is Closest to (1,-1,1).

Grade:12th pass

2 Answers

Arun
25750 Points
4 years ago
 
 we have 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
 
Hope it helps
Vikas TU
14149 Points
4 years ago
Dear student 
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 = 1
26c +2 = 1 
c = -1/26 
P = (1,-1,1) - 1/26 ( 4,3,1) 
P = (0.84 , -1.11 , 0.96) 
Hope this helps 

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