Answer
(a) $x+y+z=3$
(b) $x=y=z$
Work Step by Step
a)
Our aim is to determine the tangent plane equation.
The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1)
From the given data, we have $f(x,y,z)=(1,1,1)$
Equation (1), becomes:
Thus,$(x-1)(-2)+(y-1)(-2)+(z-1)(-2)=0$
Hence, $x+y+z=3$
b)
Our aim is to determine the normal line equation.
The general form is:
$\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ ...(2)
From the given data, we have $f(x,y,z)=(1,1,1)$
Equation (2), becomes:
Thus, $\dfrac{(x-1)}{-2}=\dfrac{(y-1)}{-2}=\dfrac{(z-1)}{-2}$
or, $x=y=z$
