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
Verified661.8k+ 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
What is BLO What is the full form of BLO class 8 social science CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
10 examples of friction in our daily life
Draw a diagram of nephron and explain its structur class 11 biology CBSE
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE
A Paragraph on Pollution in about 100-150 Words
Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE
Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?