Answer
(a) $x-2y+2z=-1$
(b) $\dfrac{(x-3)}{-1}=\dfrac{(y-1)}{2}=\dfrac{(z+1)}{-2}$
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 $f(x,y,z)=(3,1,-1)$
Equation (1), becomes:
Thus, $(x-3)(-1)+(y-1)(2)-(z+1)(2)=0$
or, $-x+3+2y-2-2z-2=0$
Hence, $x-2y+2z=-1$
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 $f(x,y,z)=(3,1,-1)$
Equation (2), becomes:
Hence, $\dfrac{(x-3)}{-1}=\dfrac{(y-1)}{2}=\dfrac{(z+1)}{-2}$
