Find the Point on the plane 4x+3y+z=2 that is Closest to (1,-1,1).
Akif , 5 Years ago
Grade 12th pass
Differential Calculus> Find the Point on the plane 4x+3y+z=2 tha...
2 Answers
Arun
Last Activity: 5 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
Last Activity: 5 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)
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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