# Question #cdb28

Apr 30, 2017

See below.

#### Explanation:

The normal vector to a surface $f \left(x , y , z\right) = 0$ is given by

$\nabla f = \left({f}_{x} , {f}_{y} , {f}_{z}\right)$

Given

${p}_{0} = \left(1 , 2 , 4\right)$ and
$p = \left(x , y , z\right)$

the tangent plane at this point is given by

$\Pi \to \left\langlep - {p}_{0} , \vec{n}\right\rangle$ where

$\left\langle\cdot , \cdot\right\rangle$ represents the scalar product of two vectors and

$\vec{n} = \nabla f \left({p}_{0}\right) = \left(2 {x}_{0} , 2 {y}_{0} , 1\right) = \left(2 , 4 , 4\right)$ so

$\left(x - 1\right) 2 + \left(y - 2\right) 4 + \left(z - 4\right) 4 = 0$

The normal line is given by

$L \to p = {p}_{0} + \lambda \vec{n}$ or

$\left\{\begin{matrix}x = 1 + 2 \lambda \\ y = 2 + 4 \lambda \\ z = 4 + 4 \lambda\end{matrix}\right.$

Apr 30, 2017

We are using the directional derivative which tells us that the normal vector is the gradient, ie :

$m a t h b f n = m a t h b f \nabla f \left(m a t h b f x\right) = < 2 x , 2 y , 1 {>}_{\left(1 , 2 , 4\right)} = < 2 , 4 , 1 >$

So, the tangent plane has $m a t h b f n$ as it's normal vector, and also, like any other plane, has equation:

$\left(m a t h b f r - m a t h b f {r}_{0}\right) \cdot m a t h b f n = 0 \implies m a t h b f r \cdot m a t h b f n = \textcolor{b l u e}{m a t h b f {r}_{0} \cdot m a t h b f n} = \textcolor{b l u e}{\alpha}$

The normal line passes through point $\left(1 , 2 , 4\right)$, so $m a t h b \mathfrak{_} o = < 1 , 2 , 4 >$ and $m a t h b f {r}_{0} \cdot m a t h b f n = < 1 , 2 , 4 > \cdot < 2 , 4 , 1 > = 14 = \alpha$

So the tangent plane is:

$2 x + 4 y + z = 14$

The normal line, $m a t h b f l$, passes through $\left(1 , 2 , 4\right)$ and has direction $< 2 , 4 , 1 >$. With $\lambda$ as the parameter:

$m a t h b f l = < 1 , 2 , 4 > + \lambda < 2 , 4 , 1 >$