The equation of a plane is defined as

where  is the normal vector of the plane. 

To find the normal vector, we first get two vectors on the plane 

 and 

and find their cross product. 

The cross product is defined as the determinant of the matrix

Which is

Which tells us the normal vector is 

Using the point  and the normal vector to find the equation of the plane yields

Simplified gives the equation of the plane 

Planes that are parallel to each other only differ (if at all) by the constant on the right-hand side (when both sides are simplified). Since has the same coefficients as the given plane, they are parallel to each other.

We can begin by rewriting the expression for the cylinder as follows

.

This tells us that .  Plugging this back into the equation for the plane  to find .

This gives us the representation of the curve of intersection as

.

The equation of a plane is defined as

where 

 is the normal vector of the plane. 

To find the normal vector, we first get two vectors on the plane 

 and 

and find their cross product. 

The cross product is defined as the determinant of the matrix

Which is

Which tells us the normal vector is 

Using the point 

 and the normal vector to find the equation of the plane yields

Simplified gives the equation of the plane 

To find the angle between the planes, we find the angle between their normal vectors.

We have

, for the first plane, and

, for the second plane.

The angle between these two vectors is

Take the cross product of the given points. This will result in an orthogonal vector to both  and , whose components will be the corresponding direction ratios.

Equation of the plane is

Now plug in  or , you will have the answer

To find the equation for the plane, we must first find two vectors contained within the plane:

We found these by subtracting the terminal point from the initial point for each vector (the terminal points being B and C, and the initial points A and B, respectively). 

Now, 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 must find the cross product of the two vectors, in order to determine the normal vector to the plane. 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:

Now, we write out the equation of a plane, given by 

, where a, b, and c are the elements of the normal vector , and  is any point on the plane.

Our normal vector is equal to , and we can pick point A, so the equation of the plane is

which becomes

This answer should make sense because none of the points have a z component, so they are all on the x-y plane, thus the only plane containing all three is the one where z has no value. 

The equation of a plane is given by:

, where a, b, and c are the elements of the normal vector , and  is any point on the plane.

Using the above formula, we get

which becomes

To find the unit tangent vector for a given curve, we must find the derivative of each of the components of the curve, and then divide this vector - the tangent vector - by the magnitude of the tangent vector itself.

To start, let's find the derivatives:

We used the following rules to find the derivatives:

, , 

Next, we must find the magnitude of the tangent vector, by taking the square root of the sum of the squares the x, y, and z components. Note that the vector we were given was in unit vector notation, and i, j, and k are each equal to one unit, so they don't impact the magnitude of the vector (they simply indicate direction).

Finally, divide the tangent vector by its magnitude:

To find the equation of a plane, we need a point on the plane and the normal vector to the plane, given by the cross product of two vectors on the plane. 

To start, let's find these two vectors:

We found these by subtracting terminal points (B, C) and initial points (A, B, respectively).

Now, 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:

The equation of a plane is given by:

, where a, b, and c are the elements of the normal vector , and  is any point on the plane.

Our normal vector is , and the point we can choose is . Using the above formula, we get