Question

How do you find the equation for the tangent plane to the surface \[{x^2} + 2{z^2} = {y^2}\] through $(1,3, - 2)$ ?

Answer 1

Hint:In this question, we are given a plane in 3-D, that is, a plane that lies in three dimensions, the equation of this plane includes x, y and z so it lies in a plane containing x, y and z-axis. The equation of the given plane can also be written as a function of x, y and z.
To find the equation of the tangent to this plane surface lying in three dimensions at a given point in vector, we use del, or gradient operator, because using del or gradient operator we differentiate the function with respect to the variable given in the denominator while keeping other variables constant.

Complete step by step answer:
We are given \[{x^2} + 2{z^2} = {y^2}\]
It can be rewritten as –
\[{x^2} - {y^2} + 2{z^2} = 0\]
This equation can be written as a function of x, y and z as follows –
$f(x,y,z) = {x^2} - {y^2} + 2{z^2}$
Del or gradient operator of this function is given as –
$
\nabla f(x,y,z) = \dfrac{{\delta f}}{{\delta x}}\hat i + \dfrac{{\delta f}}{{\delta y}}\hat j + \dfrac{{\delta
f}}{{\delta z}}\hat k \\
\Rightarrow \nabla f = \dfrac{{\delta ({x^2} - {y^2} + 2{z^2})}}{{\delta x}}\hat i + \dfrac{{\delta ({x^2}
- {y^2} + 2{z^2})}}{{\delta y}}\hat j + \dfrac{{\delta ({x^2} - {y^2} + 2{z^2})}}{{\delta z}}\hat k \\
\Rightarrow \nabla f = 2x\hat i - 2y\hat j + 4z\hat k \\
$
We have to find the tangent to the plane surface at $(1,3, - 2)$ -
$
\nabla f(1,3, - 2) = 2(1)\hat i - 2(3)\hat j + 4( - 2)\hat k \\
\Rightarrow \nabla f(1,3, - 2) = 2\hat i - 6\hat j - 8\hat k \\
$
This equation is the equation of the normal at the plane surface through $(1,3, - 2)$ , the equation of the tangent to a plane surface is given as –
\[\vec r.\vec n = \vec a.\vec n\]
For the tangent of the given plane –
$\vec r = x\hat i + y\hat j + z\hat k,\,\vec a = 1\hat i + 3\hat j - 2\hat k,\,and\,\vec n = 2\hat i - 6\hat j
- 8\hat k$
So, we get –
$
(x\hat i + y\hat j + z\hat k).(2\hat i - 6\hat j - 8\hat k) = (1\hat i + 3\hat j - 2\hat k).(2\hat i - 6\hat j -
8\hat k) \\
x \times 2 + y \times ( - 6) + z \times ( - 8) = 1 \times 2 + 3 \times ( - 6) + ( - 2) \times ( - 8) \\
\Rightarrow 2x - 6y - 8z = 2 - 18 + 16 \\
\Rightarrow 2x - 6y - 8z = 0 \\
\Rightarrow x - 3y - 4z = 0 \\
$
Hence, the equation for the tangent plane to the surface \[{x^2} + 2{z^2} = {y^2}\] through $(1,3, -
2)$ is given as $x - 3y - 4z = 0$ .

Note: While doing the dot product of two vectors, we multiply the coefficient of unit vectors $\hat i,\,\hat j\,and\,\hat k$ with each other, the signs of the coefficients are also multiplied as you can see in this question. Del or gradient operator is used while we have to differentiate a multidimensional function, that is, a function that has more than one variable in its equation.