Find the equation of the tangent plane and normal to a surface

  • #1
chwala
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Homework Statement
See attached ##Q.64## highlighted in red.
Relevant Equations
vector differentiation
1714647682038.png


In my line i have,

##\dfrac{∂r}{du} = \vec{i} +\dfrac{1}{2}u \vec{k} = \vec{i} +1.5 \vec{k}##

##\dfrac{∂r}{dv} = \vec{j} -\dfrac{1}{2}v \vec{k} = \vec{j} -0.5\vec{k}##

The normal to plane is given by,

##\dfrac{∂r}{du}× \dfrac{∂r}{dv} = -\dfrac{3}{2} \vec{ i} + \dfrac{1}{2}\vec{j}+\vec{k}##

The equation of the tangent to the plane is given by,

...

##-\dfrac{3}{2} (x-3)+ \dfrac{1}{2}(y-1)+1(z-2)=0##

##-3(x-3)+(y-1)+2(z-2)=0##

##-3x+9+y-1+2z-4=0##

##-3x+y+2z+4=0##

##3x-y-2z=4##

looks good... just need to confirm/check on the Normal line equation. Insight is welcome.

Cheers.
 
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  • #2
chwala said:
The equation of the tangent to the plane
Correction: The equation of the tangent plane; that is the plane that is tangent to the surface at the given point.
chwala said:
just need to confirm/check on the Normal line equation.
Did you calculate the equation of the normal line? If so you didn't include it in your work. You have a vector that has the same direction as the normal line, and a point on this line, so it's quick work to come up with the equation of that line.
 
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