Find the unit normal vector to the surface $4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z=40$ at the point $\left( 2,-1,2 \right)$.
Answer
Verified578.7k+ views
Hint: We use the fact that gradient of the surface at a given point is the normal vector for that surface. So, we first find the normal vector for the given surface. Once we find the normal vector, we convert the normal vector to a unit vector (vector of magnitude ‘1’) to get the required result.
Complete step by step answer:
We have the equation of the surface given as $4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z=40$. We need to find the unit normal vector at the point $\left( 2,-1,2 \right)$ for the given surface.
Let us assume the equation of the surface be $f(x,y,z)=0\Leftrightarrow 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40=0$.
We first find the normal vector for the given surface f. We know that the gradient of the surface ‘f’ at a given point $\left( x,y,z \right)$ is vector normal to the surface.
The gradient of the surface is defined as ${{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)}}={{\left. \left( \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z} \right) \right|}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)}}$.
Now we find the gradient of the surface $f(x,y,z)=4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40$.
Normal vector of the surface is ${{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z} \right) \right|}_{\left( 2,-1,2 \right)}}$.
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial x},\dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial y},\dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial z} \right) \right|}_{\left( 2,-1,2 \right)}}\].
We know that while doing partial derivative w.r.t x $\left( \dfrac{\partial }{\partial x} \right)$, we take all other terms as constant and apply the derivative only for ‘x’.
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \left( 4{{z}^{3}}-6x{{y}^{2}}z-0 \right),\left( 0-6{{x}^{2}}yz-0 \right),\left( 12x{{z}^{2}}-3{{x}^{2}}{{y}^{2}}-0 \right) \right) \right|}_{\left( 2,-1,2 \right)}}\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \left( 4{{z}^{3}}-6x{{y}^{2}}z \right),\left( -6{{x}^{2}}yz \right),\left( 12x{{z}^{2}}-3{{x}^{2}}{{y}^{2}} \right) \right) \right|}_{\left( 2,-1,2 \right)}}\].
Normal vector of the surface $f(x,y,z)$ is\[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 4{{\left( 2 \right)}^{3}}-6\left( 2 \right){{\left( -1 \right)}^{2}}\left( 2 \right) \right),\left( -6{{\left( 2 \right)}^{2}}\left( -1 \right)\left( 2 \right) \right),\left( 12\left( 2 \right){{\left( 2 \right)}^{2}}-3{{\left( 2 \right)}^{2}}{{\left( -1 \right)}^{2}} \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 4.\left( 8 \right)-6.\left( 2 \right).\left( 1 \right).\left( 2 \right) \right),\left( -6.\left( 4 \right).\left( -1 \right).\left( 2 \right) \right),\left( 12.\left( 2 \right).\left( 4 \right)-3.\left( 4 \right).\left( 1 \right) \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 32-24 \right),48,\left( 96-12 \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( 8,48,84 \right)\] ---(1).
We know that the unit vector of any given vector $\left( x,y,z \right)$ is $\pm \dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\times \left( x,y,z \right)$. Using this we find the unit normal vector for the surface $f(x,y,z)$.
So, unit normal vector of the surface $f(x,y,z)$ is $\pm \dfrac{1}{\sqrt{{{8}^{2}}+{{48}^{2}}+{{84}^{2}}}}\times \left( 8,48,84 \right)$.
Note: We can verify that the magnitude of the obtained unit normal vector is ‘1’. We can see that the normal vector and unit normal vectors are parallel to each other having different magnitudes. We should not write a normal vector as a unit normal vector as its magnitude is not ‘1’.
Complete step by step answer:
We have the equation of the surface given as $4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z=40$. We need to find the unit normal vector at the point $\left( 2,-1,2 \right)$ for the given surface.
Let us assume the equation of the surface be $f(x,y,z)=0\Leftrightarrow 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40=0$.
We first find the normal vector for the given surface f. We know that the gradient of the surface ‘f’ at a given point $\left( x,y,z \right)$ is vector normal to the surface.
The gradient of the surface is defined as ${{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)}}={{\left. \left( \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z} \right) \right|}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)}}$.
Now we find the gradient of the surface $f(x,y,z)=4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40$.
Normal vector of the surface is ${{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z} \right) \right|}_{\left( 2,-1,2 \right)}}$.
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial x},\dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial y},\dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial z} \right) \right|}_{\left( 2,-1,2 \right)}}\].
We know that while doing partial derivative w.r.t x $\left( \dfrac{\partial }{\partial x} \right)$, we take all other terms as constant and apply the derivative only for ‘x’.
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \left( 4{{z}^{3}}-6x{{y}^{2}}z-0 \right),\left( 0-6{{x}^{2}}yz-0 \right),\left( 12x{{z}^{2}}-3{{x}^{2}}{{y}^{2}}-0 \right) \right) \right|}_{\left( 2,-1,2 \right)}}\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \left( 4{{z}^{3}}-6x{{y}^{2}}z \right),\left( -6{{x}^{2}}yz \right),\left( 12x{{z}^{2}}-3{{x}^{2}}{{y}^{2}} \right) \right) \right|}_{\left( 2,-1,2 \right)}}\].
Normal vector of the surface $f(x,y,z)$ is\[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 4{{\left( 2 \right)}^{3}}-6\left( 2 \right){{\left( -1 \right)}^{2}}\left( 2 \right) \right),\left( -6{{\left( 2 \right)}^{2}}\left( -1 \right)\left( 2 \right) \right),\left( 12\left( 2 \right){{\left( 2 \right)}^{2}}-3{{\left( 2 \right)}^{2}}{{\left( -1 \right)}^{2}} \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 4.\left( 8 \right)-6.\left( 2 \right).\left( 1 \right).\left( 2 \right) \right),\left( -6.\left( 4 \right).\left( -1 \right).\left( 2 \right) \right),\left( 12.\left( 2 \right).\left( 4 \right)-3.\left( 4 \right).\left( 1 \right) \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 32-24 \right),48,\left( 96-12 \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( 8,48,84 \right)\] ---(1).
We know that the unit vector of any given vector $\left( x,y,z \right)$ is $\pm \dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\times \left( x,y,z \right)$. Using this we find the unit normal vector for the surface $f(x,y,z)$.
So, unit normal vector of the surface $f(x,y,z)$ is $\pm \dfrac{1}{\sqrt{{{8}^{2}}+{{48}^{2}}+{{84}^{2}}}}\times \left( 8,48,84 \right)$.
Note: We can verify that the magnitude of the obtained unit normal vector is ‘1’. We can see that the normal vector and unit normal vectors are parallel to each other having different magnitudes. We should not write a normal vector as a unit normal vector as its magnitude is not ‘1’.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE
State and explain Hardy Weinbergs Principle class 12 biology CBSE
Write any two methods of preparation of phenol Give class 12 chemistry CBSE
Which of the following statements is wrong a Amnion class 12 biology CBSE
Differentiate between action potential and resting class 12 biology CBSE
Trending doubts
What are the major means of transport Explain each class 12 social science CBSE
Which are the Top 10 Largest Countries of the World?
Draw a labelled sketch of the human eye class 12 physics CBSE
How much time does it take to bleed after eating p class 12 biology CBSE
Explain sex determination in humans with line diag class 12 biology CBSE
When was the first election held in India a 194748 class 12 sst CBSE