The equation of a plane is given by

where the normal vector to the plane is  and a point on the plane . 

The normal vector is given by the cross product of the two vectors in the plane.

First, we can write the determinant in order to take the cross product of the two vectors:

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

Plugging all of our known information into the equation above, we get

which simplifies to 

To find the equation of the plane that contains a point  and has a normal vector , you use the formula:

Using the information from the problem statement, we get

Simplifying, we get

To find the equation of the plane that contains a point  and has a normal vector , you use the formula:

Using the information from the problem statement, we get

Simplifying, we get

To find the equation of the plane that contains a point  and has a normal vector , you use the formula:

Using the information from the problem statement, we get

Simplifying, we get

To find the equation of any plane, we need a point on the plane and its normal vector. The normal vector of this plane is given by the equation of the parallel plane, due to the fact that they have the same normal vector. Using the point on the plane  and the normal vector , we use the formula

Plugging in what we know, we get

Simplifying, we get 

First, we need to form two vectors on the plane to get the normal vector to the plane. This is done from the following operation:

We then take the cross product of these vectors, which gets us the normal vector

Plugging the point  and the normal vector into the equation of the plane, we get

Simplifying, we get

Find the equation of the plane tangent to the surface  at the point where  and . 

                                                                                 (1) 

The equation of the plane tangent to the surface defined by  is given by the formula: 

              (2)

In Equation (2)  and  are the partial derivatives of  with respect to  and , respectively. For this particular problem we have  

Let's fill in Equation (2) term-by-term: 

Compute the partial derivative and then evaluate both at  

- Partial with respect to x 

- Partial with respect to 

Now fill in Equation (2) and simplify to get the equation of the tangent plane: 

Therefore the equation of the tangent plane to the surface  at the point  is simply 

To find the equation of the plane, we use the formula , where the point given is  and the normal vector . Plugging in what we were given in the problem statement, we get . Manipulating the equation through algebra to isolate the variables, we get  .

First we need to find the tangent vector, and find its magnitude.

Now we can set up our arc length integral

First we need to find the tangent vector, and find its magnitude.

Now we can set up our arc length integral