Question

How do you find a unit vector normal to the surface \[{x^3} + {y^3} + 3xyz = 3\] at the point\[\left( {1,2, - 1} \right)?\]

Answer 1

Hint:This question involves the operation of addition/ subtraction/ multiplication/division. We need to know the formula for finding the unit normal vector perpendicular to the surface. Also, we need to know how to find gradient values. We need to know how to substitute the point with the given equation in the question.

Complete step by step solution:
The given equation is shown below,
\[{x^3} + {y^3} + 3xyz = 3\]at point\[\left( {1,2, - 1} \right)\].
The above equation can also be written as,
\[f\left( {x,y,z} \right) = {x^3} + {y^3} + 3xyz - 3 \to \left( 1 \right)\]
The formula for finding unit vector normal to the surface is given below,
Unit normal vector perpendicular to the surface\[ = \dfrac{{\nabla F}}{{\left| {\nabla F} \right|}}
\to \left( 2 \right)\]
Here \[F\]can also be written as\[f\left( {x,y,z} \right)\]. Let’s find the\[\nabla F\]value.
The formula for finding the value \[\nabla F\]is given below,
\[\nabla F = \left( {\overrightarrow i \dfrac{\partial }{{\partial x}} + \overrightarrow j
\dfrac{\partial }{{\partial y}} + \overrightarrow k \dfrac{\partial }{{\partial z}}} \right)\left( {f\left(
{x,y,z} \right)} \right)\]
So, we get
\[\nabla F = \left( {\overrightarrow i \dfrac{\partial }{{\partial x}} + \overrightarrow j
\dfrac{\partial }{{\partial y}} + \overrightarrow k \dfrac{\partial }{{\partial z}}} \right)\left( {{x^3}
+ {y^3} + 3xyz - 3} \right)\]
\[\nabla F = \overrightarrow i \dfrac{\partial }{{\partial x}}\left( {{x^3} + {y^3} + 3xyz - 3} \right) +
\overrightarrow j \dfrac{\partial }{{\partial y}}\left( {{x^3} + {y^3} + 3xyz - 3} \right) +
\overrightarrow k \dfrac{\partial }{{\partial z}}\left( {{x^3} + {y^3} + 3xyz - 3} \right)\]\[ \to \left( 3
\right)\]
We know that,
\[\dfrac{{\partial {x^n}}}{{\partial x}} = n{x^{n - 1}}\]
We can make the easy calculation by using this formula as follows,
\[\dfrac{\partial }{{\partial x}}\left( {{x^3} + {y^3} + 3xyz - 3} \right) = 3{x^2} + 3yz\]
(Here\[{y^3}\]and\[ - 3\] are constant so if we differentiate these value the answer become zero)
\[\dfrac{\partial }{{\partial y}}\left( {{x^3} + {y^3} + 3xyz - 3} \right) = 3{y^2} + 3xz\]And
\[\dfrac{\partial }{{\partial z}}\left( {{x^3} + {y^3} + 3xyz - 3} \right) = 3xy\]
So, the equation\[\left( 3 \right)\]becomes,
\[
\left( 3 \right) \to \nabla F = \overrightarrow i \dfrac{\partial }{{\partial x}}\left( {{x^3} + {y^3} +
3xyz - 3} \right) + \overrightarrow j \dfrac{\partial }{{\partial y}}\left( {{x^3} + {y^3} + 3xyz - 3}
\right) + \overrightarrow k \dfrac{\partial }{{\partial z}}\left( {{x^3} + {y^3} + 3xyz - 3} \right)
\\
\\
\]
\[\nabla F = \overrightarrow i \left( {3{x^2} + 3yz} \right) + \overrightarrow j \left( {3{y^2} + 3xz}
\right) + \overrightarrow k \left( {3xy} \right)\]
The point is\[\left( {x,y,z} \right)\], so let’s replace them with\[\left( {1,2, - 1} \right)\].
So, we get
\[
\nabla F = \overrightarrow i \left( {3{{\left( 1 \right)}^2} + 3 \times 2 \times - 1} \right) +
\overrightarrow j \left( {3{{\left( 2 \right)}^2} + 3 \times 1 \times - 1} \right) + \overrightarrow k
\left( {3 \times 1 \times 2} \right) \\
\nabla F = \overrightarrow i \left( {3 - 6} \right) + \overrightarrow j \left( {12 - 3} \right) +
\overrightarrow k \left( 6 \right) \\
\nabla F = - 3\overrightarrow i + 9\overrightarrow j + 6\overrightarrow k \\
\]
So, the equation\[\left( 2 \right)\]becomes,
\[
\dfrac{{\nabla F}}{{\left| {\nabla F} \right|}} = \dfrac{{ - 3\overrightarrow i + 9\overrightarrow j
+ 6\overrightarrow k }}{{\sqrt {{{\left( { - 3} \right)}^2} + {9^2} + {6^2}} }} \\
\dfrac{{\nabla F}}{{\left| {\nabla F} \right|}} = \dfrac{{ - 3\overrightarrow i + 9\overrightarrow j
+ 6\overrightarrow k }}{{\sqrt {9 + 81 + 36} }} \\
\dfrac{{\nabla F}}{{\left| {\nabla F} \right|}} = \dfrac{{ - 3\overrightarrow i + 9\overrightarrow j
+ 6\overrightarrow k }}{{\sqrt {126} }} \\
\dfrac{{\nabla F}}{{\left| {\nabla F} \right|}} = \dfrac{{ - 3\overrightarrow i + 9\overrightarrow j
+ 6\overrightarrow k }}{{3\sqrt {14} }} \\
\dfrac{{\nabla F}}{{\left| {\nabla F} \right|}} = \dfrac{{ - \overrightarrow i + 3\overrightarrow j +
2\overrightarrow k }}{{\sqrt {14} }} \\
\]
(Here, note that the value of\[{\overrightarrow i ^2}\]is equal to\[1\], as same as
for\[\overrightarrow k \]and\[\overrightarrow j \].
So, the final answer is,
The value of unit vector normal to the surface\[{x^3} + {y^3} + 3xyz = 3\]at a point\[\left( {1,2, - 1}
\right)\]is,
\[\dfrac{{\nabla F}}{{\left| {\nabla F} \right|}} = \dfrac{{ - \overrightarrow i + 3\overrightarrow j
+ 2\overrightarrow k }}{{\sqrt {14} }}\]

Note: This question involves the operation of addition/ subtraction/ multiplication/ division. We would remember the basic differentiation process and its formulae. Also, note the formula for finding the unit normal vector perpendicular to the surface. Note that the determinant can be eliminated with the sum of squares in the square root. When more than two variables are used, then we would use\[\partial \]the function.