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:
The answer would be 54 words per minute
Step-by-step explanation:
Answer:
(-1,2)
Step-by-step explanation:
2(3x+5)+3x=1
6x+10+3x=1
9x+10=1
9x=-9
x=-1
y=3(-1)+5
y=-3+5
y=-2
We have the equation f(x)=6x-5 to start with
Now in order to answer the question we have to know that f(5) is the same thing as setting x=5 and solving for f(5)
(A) f(5) = 6(5) - 5 = 30 - 5 = 25
(B) f(-3) = 6(-3) - 5 = (-18) - 5 = -23
So f(5) = 25 and f(-3) = -23
