Answer
(a) $x+2y+6z=12$
(b) $(x-2)=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{6}$
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 $f(x,y,z)=(2,2,1)$
Equation (1), becomes:
Thus, $(x-2)(4)+(y-2)(8)+(z-1)(24)=0$
or, $4x-8+8y-16+24z-24=0$
Hence, $4x+8y+24z=48$ or, $x+2y+6z=12$
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 $f(x,y,z)=(2,2,1)$
Equation (2), becomes:
Thus, $\dfrac{(x-2)}{4}=\dfrac{(y-2)}{8}=\dfrac{(z-1)}{24}$
Hence, $(x-2)=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{6}$
