Question

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`?

Answer 1

See below.

Explanation:

Calling
`Sigma->f(x,y,z)=y^2 + 3 x^2 + z^2 - 4=0`
and considering `p = (x,y,z)` such that
`p in Sigma`, we have

`vec n = (p-p_1) xx (p-p_2)` 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 `grad f = ((partial f)/(partial x), (partial f)/(partial y), (partial f)/(partial z)) =2(3x,y,z)`
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) = (0.531359, 1.59408, -0.782233)` and
`p_(t_2) = (0.242834, 0.728503, 1.81449)` the tangency points

and also

`(grad _1 f)/norm(grad_1 f) =(vec n_1)/norm(vec n_1) = (0.668034, 0.668034, -0.327812)`
`(grad_2 f)/norm(grad_2 f) =(vec n_2)/norm(vec n_2)=-(0.349138, 0.349138, 0.869601)`

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 `(cdot xx cdot)` represents the vector product and `<< cdot, cdot >>` the scalar product. Also `norm(cdot)` represents the vector norm.