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

[chwala]

Homework Statement

See attached $Q.64$ highlighted in red.

Relevant Equations

vector differentiation

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.

 

[Mark44]

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.

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.