Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point.
x - z = 4arctan(yz) , (1 + pi, 1, 1)
My work-
x - z - 4arctan(yz)
<d/dx, d/dy, d/dz> = <1, -(4z)/(y^2z^2+1), -(4y)/(y^2z^2+1) - 1 )
a)
F(1+pi, 1, 1)= <1, -2, -3> is the normal vector
so the equation would be- 1(x-(1+pi))-2(y-1)-3(z-1)
b)
parametric equations- x= (1+pi)+t , y= 1-2t, z= 1-3t
symetric equations- (x-(1+pi))/1 , (y-1)/(-2), (z-1)/(-3)
is this correct? I wasn't sure if i should do:
x-z-4arctan(yz)
or
(x-z)/(arctan(yz))=4
x - z = 4arctan(yz) , (1 + pi, 1, 1)
My work-
x - z - 4arctan(yz)
<d/dx, d/dy, d/dz> = <1, -(4z)/(y^2z^2+1), -(4y)/(y^2z^2+1) - 1 )
a)
F(1+pi, 1, 1)= <1, -2, -3> is the normal vector
so the equation would be- 1(x-(1+pi))-2(y-1)-3(z-1)
b)
parametric equations- x= (1+pi)+t , y= 1-2t, z= 1-3t
symetric equations- (x-(1+pi))/1 , (y-1)/(-2), (z-1)/(-3)
is this correct? I wasn't sure if i should do:
x-z-4arctan(yz)
or
(x-z)/(arctan(yz))=4
