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.
Hello!
A cylinder has 3 faces 2 edges and 0 vertices
The answer is cylinder
Hope this helps!
Answer:
No solution
Step-by-step explanation:
I used this online calculator to find the answer. It shows the steps and everything. https://www.symbolab.com/solver?or=gms&query=2x%2B3y%3D62x%2B3y%3D7
Answer: -30
Step-by-step explanation:
(-6) x (7) - (6) x (-2)
(-42) - (-12)=
-30
Hope this helps :)
Answer:
42×15=x.....................
Step-by-step explanation:
:)
