Answer:
plz if you knew if you new it plz tell me
Given plane Π : f(x,y,z) = 4x+3y-z = -1
Need to find point P on Π that is closest to the origin O=(0,0,0).
Solution:
First step: check if O is on the plane Π : f(0,0,0)=0 ≠ -1 => O is not on Π
Next:
We know that the required point must lie on the normal vector <4,3,-1> passing through the origin, i.e.
P=(0,0,0)+k<4,3,-1> = (4k,3k,-k)
For P to lie on plane Π , it must satisfy
4(4k)+3(3k)-(-k)=-1
Solving for k
k=-1/26
=>
Point P is (4k,3k,-k) = (-4/26, -3/26, 1/26) = (-2/13, -3/26, 1/26)
because P is on the normal vector originating from the origin, and it satisfies the equation of plane Π
Answer: P(-2/13, -3/26, 1/26) is the point on Π closest to the origin.
First plug in (x+h) for x in the function.
f(x+h)= 2(x-h)^2-3(x-h) = 2(x^2-2xh+h^2)-3x-3h =
2x^2-4xh+2h^2-3x-3h - 2x^2 +3x =
(-4xh +2h^2-3h)/h
-4x +2h-3
In the first number, 2 is in the thousand's place, it means two thousand.
in the second number, 2 is in the hundred's place. It means two hundred.
