# Given the surface f(x,y,z)=y^2 + 3 x^2 + z^2 - 4=0 and the points p_1=(2,1,1) and p_2=(3,0,1) determine the tangent plane to f(x,y,z)=0 containing the points p_1 and p_2?

Dec 27, 2016

See below.

#### Explanation:

Calling
$\Sigma \to f \left(x , y , z\right) = {y}^{2} + 3 {x}^{2} + {z}^{2} - 4 = 0$
and considering $p = \left(x , y , z\right)$ such that
$p \in \Sigma$, we have

$\vec{n} = \left(p - {p}_{1}\right) \times \left(p - {p}_{2}\right)$ is a vector normal to the plane
$\Pi$ defined by the points $p , {p}_{1} , {p}_{2}$

Now, the vector $\vec{n}$ can be computed over $\Sigma$ as $\nabla f = \left(\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}\right) = 2 \left(3 x , y , z\right)$
The $\Sigma$ tangent point is then determined by the equations

{ (n_x/norm(vec n)=pmf_x/norm(grad f)), (<< grad f, p_1-p_2 >> = 0), (f(x,y,z)=0) :}(1)

or

{((1 - z)/sqrt[(x + y-3)^2 + 2 (z-1)^2] pm (3 x)/sqrt[ 9 x^2 + y^2 + z^2]=0),(y-3x=0),(y^2 + 3 x^2 + z^2 - 4=0):}(2)

Solving for $x , y , z$ we obtain

${p}_{{t}_{1}} = \left(0.531359 , 1.59408 , - 0.782233\right)$ and
${p}_{{t}_{2}} = \left(0.242834 , 0.728503 , 1.81449\right)$ the tangency points

and also

$\frac{{\nabla}_{1} f}{\left\lVert {\nabla}_{1} f \right\rVert} = \frac{{\vec{n}}_{1}}{\left\lVert {\vec{n}}_{1} \right\rVert} = \left(0.668034 , 0.668034 , - 0.327812\right)$
$\frac{{\nabla}_{2} f}{\left\lVert {\nabla}_{2} f \right\rVert} = \frac{{\vec{n}}_{2}}{\left\lVert {\vec{n}}_{2} \right\rVert} = - \left(0.349138 , 0.349138 , 0.869601\right)$

the corresponding normal surface vectors.

Note:

a) In (1,2) we consider only a vector component. The choice is one of

(((1 - z)/sqrt[(x + y-3)^2 + 2 (z-1)^2] - (3 x)/sqrt[ 9 x^2 + y^2 + z^2]),((1 - z)/sqrt[(x + y-3)^2 + 2 (z-1)^2] - y/sqrt[ 9 x^2 + y^2 + z^2]),((x + y-3)/sqrt[(x + y-3)^2 + 2 (z-1)^2] - z/sqrt[ 9 x^2 + y^2 + z^2]))

b) The sign $\pm$ before the gradient is to qualify the two solutions to the problem.

c) Here $\left(\cdot \times \cdot\right)$ represents the vector product and $\left\langle\cdot , \cdot\right\rangle$ the scalar product. Also $\left\lVert \cdot \right\rVert$ represents the vector norm.