## Calculus 8th Edition

(a) $x+y+z=11$ (b) $(x-3)=(y-3)=(z-5)$
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 $f(x,y,z)=(3,3,5)$ Equation (1), becomes: $(x-3)(4)+(y-3)(4)+(z-5)(4)=0$ or, $4x-12+4y-12+4z-20=0$ Hence, $x+y+z=11$ 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 $f(x,y,z)=(3,3,5)$ Equation (2), becomes: $\dfrac{(x-3)}{4}=\dfrac{(y-3)}{4}=\dfrac{(z-5)}{4}$ Hence, $(x-3)=(y-3)=(z-5)$