Answer:
no
Step-by-step explanation:
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.
Answer:
A y= -3/4+6
Step-by-step explanation
I hope this helps good luck
I'm 100% sure
Good luck god bless you pass have a nice day
Answer:
ok and thank you for point
Step-by-step explanation:
a). A = {x ∈ R I 5x-8 < 7}
5x - 8 < 7 <=> 5x < 8+7 <=> 5x < 15 =>
x < 3 => A = (-∞ ; 3)
A ∩ N = {0 ; 1 ; 2}
A - N* = (-∞ ; 3) - {1 ; 2}
b). A = { x ∈ R I 7x+2 ≤ 9}
7x+2 ≤ 9 <=> 7x ≤ 7 => x ≤ 1 => x ∈ (-∞ ; 1]
A ∩ N = {0 ; 1}
A-N* = (-∞ ; 1)
c). A = { x ∈ R I I 2x-1 I < 5}
I 2x-1 I < 5 <=> -5 ≤ 2x-1 ≤ 5 <=>
-4 ≤ 2x ≤ 6 <=> -2 ≤ x ≤ 3 => x ∈ [-2 ; 3]
A ∩ N = {0 ; 1 ; 2 ; 3}
A - N* = [-2 ; 3) - {1 ; 2}
d). A = {x ∈ R I I 6-3x I ≤ 9}
I 6-3x I ≤ 9 <=> -9 ≤ 6-3x ≤ 9 <=>
-15 ≤ -3x ≤ 3 <=> -5 ≤ -x ≤ 3 =>
-3 ≤ x ≤ 5 => x ∈ [-3 ; 5]
A ∩ N = {0 ; 1 ; 2 ; 3 ; 4 ; 5}
A - N* = [-3 ; 5) - {1 ; 2 ; 3 ; 4}
