Question

Find the equation of the tangent plane and normal line to the surface $2{x^2} + {y^2} + 2z = 3$ at the point $(2, 1, -3)$

Answer 1

to solve this problem, we need to assume the surface to be the function in $x,y,z$. Then we have to differentiate the function with respect to $x,y,z$ and find the respective values at the point $(2,1, - 3)$. At last we have to write the equation for the tangent and the normal plane.

Complete step-by-step answer:
Let us assume that
$\phi (x,y,z) = 2{x^2} + {y^2} + 2z - 3$
We choose this function because $\phi = 0$, we get the surface as given in the question. Now we have to find the tangent plane and the normal line, so for that we need to find
$\dfrac{{d\phi }}{{dx}},\dfrac{{d\phi }}{{dy}},\dfrac{{d\phi }}{{dz}}$
This is because the formula for the tangent plane and the normal line involve their use.
Formula for the tangent plane
$(x - h)\dfrac{{d\phi }}{{dx}} + (y - k)\dfrac{{d\phi }}{{dy}} + (z - t)\dfrac{{d\phi }}{{dz}} = 0$
where $h,k,t$ are the coordinates of the point from which the tangent plane will pass.
Formula for the normal line is
$\dfrac{{x - h}}{{\dfrac{{d\phi }}{{dx}}}} = \dfrac{{y - k}}{{\dfrac{{d\phi }}{{dy}}}} = \dfrac{{z - t}}{{\dfrac{{d\phi }}{{dz}}}}$
Differentiation $\phi$ with respect to $x$, we get
$\dfrac{{\partial \phi }}{{\partial x}} = 4x$
Substituting $x = 2$ from the point $(2,1, - 3)$as it will pass through it.
$\dfrac{{\partial \phi }}{{\partial x}} = 8$
Differentiation $\phi$ with respect to $y$, we get
$\dfrac{{\partial \phi }}{{\partial y}} = 2y$
Substituting $y = 1$ from the point $(2,1, - 3)$as it will pass through it.
$\dfrac{{\partial \phi }}{{\partial y}} = 2$
Differentiation $\phi$ with respect to $z$and substitute$z = - 3$, we get
$\dfrac{{\partial \phi }}{{\partial z}} = 2$
Therefore equation of the tangent plan becomes
$(x - 2)8 + (y - 1)2 + (z + 3)2 = 0$
$(x - 2)4 + (y - 1) + (z + 3) = 0$
$4x + y + z = 6$
Equation of the normal becomes
$\dfrac{{x - 2}}{8} = \dfrac{{y - 1}}{2} = \dfrac{{z + 3}}{2} \Rightarrow \dfrac{{x - 2}}{4} = \dfrac{{y - 1}}{1} = \dfrac{{z + 3}}{1}$

Note: The key point to solve such geometrical $3 - D$ questions in remembering the formula for every tangent, normal, etc. as you can see in the above question, you cannot go ahead if you do not know the formula for the tangent plane and the normal line.