## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 999: 63

#### Answer

$x=-1-10t,y=1-16t,z=2-12t$

#### Work Step by Step

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 $f(x,y,z)=(-1,1,2)$ Equation (1), can be written as: $\nabla P \times \nabla Q=\lt -2,2,-1 \gt \times \lt -8,2,4 \gt=\lt 10, 16, 12 \gt$ After simplifications, we get $\dfrac{(x+1)}{5}=\dfrac{(y-1}{8}=\dfrac{(z-2)}{6}$ Therefore, the desired parametric line of equations are $x=-1-10t,y=1-16t,z=2-12t$

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