Question

Find an equation for the tangent plane and parametric equations for the normal line to the surface at the point P. x2 – xyz = 228; P(-6,8,4) Equation for the tangent plane: Edit Parametric equations for the normal line to the surface at the point P: Edit Edit z = 4 + 481

Answer 1

Given surface

228 zhr - z N

2 Let F(x, y, z) = xyz 228 = 0

Calculating the gradient vector of given surface

VF(x, y, z) =< Fr, Fy, F. >

Calculating Fr, Fy, F.

д F. - (х2 – туг — 228) = 2х — уг де

Fy (х? – туг — 228) = () — тz = 0 — 72 = -12 ду

F a (22 – xyz – 228) = 0 – xy = -cy az

=> VF(x, y, z) =< 2.0 - y2, -22, -ry

$\bigtriangledown F(x,y,z)\text{ is always normal to the surface.}$

$\text{Normal vector for the tangent plane at point (-6, 8, 4) } = \bigtriangledown F(-6, 8, 4)$

Equation of tangent plane at point (-6, 8, 4) will be given by

=> -44.0 264 + 24y - 192 + 482 – 192 = 0

=> -44.6 + 24y + 482 264 + 192 + 192 = 648

=> -11.2 + 6y + 122 162

Normal line at point P(-6,8,4) will be parallel to vector y F(-6, 8, 4)

=> Normal line at point P(-6,8,4) is parallel to <-44, 24, 48 >

Parametric equation of Normal line at point P(-6,8,4)

x = -6 + (-44)t = -6 - 444

y = 8 + 24t

2 = 4+ 487