Answer
(a) $3x+2y+3z=10$
(b) $\dfrac{(x-1)}{3}=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{3}$
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 $f(x,y,z)=(1,2,1)$
Equation (1), becomes:
Thus,$(x-1)(3)+(y-2)(2)+(z-1)(3)=0$
or, $3x-3+2y-2+3z-3=0$
$5(3x-3+2y-2+3z-3)=0$
Hence, $3x+2y+3z=10$
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 $f(x,y,z)=(1,2,1)$
Equation (2), becomes:
Hence,
$\dfrac{(x-1)}{3}=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{3}$
