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Question 7 If g is differentiable at $(l, m, n)$, then the normal line to the surface given by $g(x, y, z) = 0$ at $(l, m, n)$ can be expressed as $x = l + g_x(l, m, n)t$, $y = m + g_y(l, m, n)t$, $z = n + g_z(l, m, n)t$, $t \in \mathbb{R}$ $\frac{x - l}{g_x(l, m, n)} = \frac{y - m}{g_y(l, m, n)} = \frac{z - n}{g_z(l, m, n)}$, if denominators $\neq 0$ $\nabla g(l, m, n) \cdot (x - l, y - m, z - n) = 0$ $\nabla g(l, m, n) \times (x - l, y - m, z - n) = \vec{0}$ $\nabla g(l, m, n) \times (x - l, y - m, z - n) = 0$ $g_x(l, m, n)(x - l) + g_y(l, m, n)(y - m) + g_z(l, m, n)(z - n) = 0$ 2 pts

Name: question 7 if g is differentiable at l m n then the normal line to the surface given by gx y z 0 at l m n can be expressed as x l gxl m nt y m gyl m nt z n gzl m nt t in mathbbr fracx lg 35873
Uploaded: 2023-10-05T15:38:13-08:00
Duration: 1 min 29 s
Channel: Sri K
Description: question 7 if g is differentiable at l m n then the normal line to the surface given by gx y z 0 at l m n can be expressed as x l gxl m nt y m gyl m nt z n gzl m nt t in mathbbr fracx lg 35873

          Question 7
If g is differentiable at $(l, m, n)$, then the normal line to the surface given by $g(x, y, z) = 0$ at $(l, m, n)$ can be expressed as
$x = l + g_x(l, m, n)t$, $y = m + g_y(l, m, n)t$, $z = n + g_z(l, m, n)t$, $t \in \mathbb{R}$
$\frac{x - l}{g_x(l, m, n)} = \frac{y - m}{g_y(l, m, n)} = \frac{z - n}{g_z(l, m, n)}$, if denominators $\neq 0$
$\nabla g(l, m, n) \cdot (x - l, y - m, z - n) = 0$
$\nabla g(l, m, n) \times (x - l, y - m, z - n) = \vec{0}$
$\nabla g(l, m, n) \times (x - l, y - m, z - n) = 0$
$g_x(l, m, n)(x - l) + g_y(l, m, n)(y - m) + g_z(l, m, n)(z - n) = 0$
2 pts

$question 7 if g is differentiable at l m n then the normal line to the surface given by gx y z 0 at l m n can be expressed as x l gxl m nt y m gyl m nt z n gzl m nt t in mathbbr fracx lg 35873$

Added by Matthew J.

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