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Question: How do you find the equations for the tangent plane to the surface \[z = {x^2} - 2xy + {y^2}\] throu...

How do you find the equations for the tangent plane to the surface z=x2−2xy+y2z = {x^2} - 2xy + {y^2} through(1,2,1)\left( {1,2,1} \right) ?

Explanation

Solution

Here in this question, we have to find the equation for the tangent plane of a given surface. First rewrite the equation of surface in the form of f(x,y,z)=0f(x,y,z) = 0, then find the normal vector ∇f(x,y,z)=∂f∂xi^+∂f∂yj^+∂f∂zk^\nabla f(x,y,z) = \dfrac{{\partial f}}{{\partial x}}\widehat i + \dfrac{{\partial f}}{{\partial y}}\widehat j + \dfrac{{\partial f}}{{\partial z}}\widehat k at the given point and next find the equation of tangent plane using the vector equation of the form r→⋅n→=a→⋅n→\overrightarrow r \cdot \overrightarrow n = \overrightarrow a \cdot \overrightarrow n on simplification we get the required equation.

Complete step by step solution:
Let z=f(x,y)z = f(x,y) be the equation of a surface SS in R3{\mathbb{R}^3}, and let P=(a,b,c)P = (a,b,c) be a point on SS. Let TT be a plane which contains the point PP, and let Q=(x,y,z)Q = (x,y,z) represent a generic point on the surface SS. If the (acute) angle between the vector PQ→\overrightarrow {PQ} and the plane TT approaches zero as the point QQ approaches PP along the surface SS, then we call TT the tangent plane to SS at PP.

Consider the given equation of surface
z=x2−2xy+y2z = {x^2} - 2xy + {y^2}
rearrange the equation of the surface into the form f(x,y,z)=0f(x,y,z) = 0
∴  x2−2xy+y2−z=0\therefore \,\,{x^2} - 2xy + {y^2} - z = 0
We define the surface function as
f(x,y,z)=x2−2xy+y2−zf(x,y,z) = {x^2} - 2xy + {y^2} - z------(1)
Now, find the normal at any particular point in vector space using
∇f(x,y,z)=∂f∂xi^+∂f∂yj^+∂f∂zk^\nabla f(x,y,z) = \dfrac{{\partial f}}{{\partial x}}\widehat i + \dfrac{{\partial f}}{{\partial y}}\widehat j + \dfrac{{\partial f}}{{\partial z}}\widehat k------(2)
Substitute equation (1) in (2), then
∇f(x,y,z)=∂∂x(x2−2xy+y2−z)i^+∂∂y(x2−2xy+y2−z)j^+∂∂z(x2−2xy+y2−z)k^\nabla f(x,y,z) = \dfrac{\partial }{{\partial x}}\left( {{x^2} - 2xy + {y^2} - z} \right)\widehat i + \dfrac{\partial }{{\partial y}}\left( {{x^2} - 2xy + {y^2} - z} \right)\widehat j + \dfrac{\partial }{{\partial z}}\left( {{x^2} - 2xy + {y^2} - z} \right)\widehat k

As we know, when partially differentiating we differentiate w.r.t the variable in question while treating the other variables as constant, then
∇f(x,y,z)=(2x−2y)i^+(−2x+2y)j^+(−1)k^\nabla f(x,y,z) = \left( {2x - 2y} \right)\widehat i + \left( { - 2x + 2y} \right)\widehat j + \left( { - 1} \right)\widehat k-----(3)
Given the particular point (1,2,1)\left( {1,2,1} \right) the normal vector to the surface is given by:
Where x=1x = 1, y=2y = 2 and z=1z = 1.
Substitute the x,y,zx,y,z value in equation (3)
∇f(x,y,z)=(2(1)−2(2))i^+(−2(1)+2(2))j^+(−1)k^\nabla f(x,y,z) = \left( {2\left( 1 \right) - 2\left( 2 \right)} \right)\widehat i + \left( { - 2\left( 1 \right) + 2\left( 2 \right)} \right)\widehat j + \left( { - 1} \right)\widehat k
⇒∇f(x,y,z)=(2−4)i^+(−2+4)j^+(−1)k^\Rightarrow\nabla f(x,y,z) = \left( {2 - 4} \right)\widehat i + \left( { - 2 + 4} \right)\widehat j + \left( { - 1} \right)\widehat k
On simplification, we get
∇f(x,y,z)=−2i^+3j^−k^\nabla f(x,y,z) = - 2\widehat i + 3\widehat j - \widehat k

The tangent plane to the surface z=x2−2xy+y2z = {x^2} - 2xy + {y^2} has this normal vector and it also passes through the point (1,2,1)\left( {1,2,1} \right). It will therefore have a vector equation of the form:
r→⋅n→=a→⋅n→\overrightarrow r \cdot \overrightarrow n = \overrightarrow a \cdot \overrightarrow n -------(4)
Where r→=⟨x,y,z⟩\overrightarrow r = \left\langle {x,y,z} \right\rangle and n→=⟨−2,2,−1⟩\overrightarrow n = \left\langle { - 2,2, - 1} \right\rangle is the normal vector and a is any point in the plane i.e., a→=⟨1,2,1⟩\overrightarrow a = \left\langle {1,2,1} \right\rangle
Substitute r→\overrightarrow r ,n→\overrightarrow n and a→\overrightarrow a in the equation (4), then
⇒  ⟨x,y,z⟩⋅⟨−2,2,−1⟩=⟨1,2,1⟩⋅⟨−2,2,−1⟩\Rightarrow \,\,\left\langle {x,y,z} \right\rangle \cdot \left\langle { - 2,2, - 1} \right\rangle = \left\langle {1,2,1} \right\rangle \cdot \left\langle { - 2,2, - 1} \right\rangle
we know from the scalar dot product
−2x+2y−z=−2+4−1- 2x + 2y - z = - 2 + 4 - 1
On simplification., we get
−2x+2y−z=1- 2x + 2y - z = 1
Multiplying the whole equation by -1, then
∴  2x−2y+z=−1\therefore\,\,2x - 2y + z = - 1

Hence, the equation for the tangent plane to the surface z=x2−2xy+y2z = {x^2} - 2xy + {y^2} through the point (1,2,1)\left( {1,2,1} \right) is 2x−2y+z=−12x - 2y + z = - 1.

Note: Here we differentiate partially the function with respect to x, y and z. We treat other variables as constant while differentiating with respect to one variable. We must know about the vector equation of general form and then we can determine the equation for the tangent.