Question

Find the equation of the tangent plane and normal to the surface xyz= 4 at the point (1,2,2)

Answer 1

Answer:

The equation of tangent plane = 2x + y + z = 6

The equation of normal line $=\frac{(x-1)}{2} = \frac{(y-2)}{1} = \frac{(z-2)}{1}$

Step-by-step explanation:

As per the question,

The equation of surface is

F(x, y, z) = xyz – 4 = 0

Now,

On Differentiating this equation partially w.r.t. x, y and z respectively, we get

$\frac{\delta F}{\delta x }=yz$

$\frac{\delta F}{\delta y }= xz$

$\frac{\delta F}{\delta z }= xy$

Hence, at the point (1, 2, 2) we have

$\frac{\delta F}{\delta x }=yz=4$

$\frac{\delta F}{\delta y }= xz=2$

$\frac{\delta F}{\delta z }= xy=2$

Hence,

The equation of tangent plane at (1, 2, 2) is

(x-1) × 4 + (y-2) × 2 + (z-2) × 2 = 0

So,

2x + y + z = 6

Equation of normal line at point (1, 2, 2) is

$\frac{(x-1)}{4} = \frac{(y-2)}{2} = \frac{(z-2)}{2}$

Or,

$\frac{(x-1)}{2} = \frac{(y-2)}{1} = \frac{(z-2)}{1}$