Answer:
the intersection point is ( -5/4, -5/4, -9/8)
Step-by-step explanation:
F(x,y,z) = x² +y² -z= 0
Then find differential of each terms then we have
∀f(x)=2x
∀f(y)=2y
∀f(z)=-1
The partial differential which is the director vector is at F(x,y,z)= (1,1,2)
Vf(1,1,2) = (2,2,-1)
But the given point is ( 1.1,2)
Then the parametric equation of normal line will be
x= 1+2t
y= 1+2t
z= 2 -t
The parametric equation can be formed as
(1+2t)² + (1+2t)² - (2-t)= 0
If we expand the expression above we have,
2(1+4t+4t²) - 2+t= 0
0= 8t² + 9t
t= 0 or t= -9/8
Substitute the value of t into the parametric equation above
At t=0
x= 1+2t; x= 1+2(0)=1
y= 1+2t; y= 1+2(0)=1
z=2 -t ; z= 2-(0)= 2
At =0 , we have (1,1,2)
At t= -9/8
x= 1+2t; x= 1+2(-9/8)=-5/4
y= 1+2t; y= 1+2(-9/8)=-5/4
z=2 -t ; z= 2-(-9/8)= -9/8
Therefore, the intersection point is ( -5/4, -5/4, -9/8)
