To find the equation of the plane given by three points, we need the normal vector to the plane and a point on the plane.

We first must find two vectors on the plane:

These were found by taking the difference between the first and second, and second and third points 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:

Now, we plug this, and a point of our choosing, into the equation for a plane:

, where  is the normal vector and  is any given point.

Plugging this in, we get

Using the formula for a plane , where the point given is  and the normal vector is . Plugging in the known values, you get . Manipulating this equation through algebra gives you the answer 

To find the equation of a plane, we need a point on the plane and the normal vector to the plane. 

First, we must find two vectors on the plane using the given points (taking the difference between the two points in each vector):

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:

Now, we choose a point (for example, ) and use the normal vector given by  from the cross product:

Plugging all of this in, and simplifying, we get

The formula of a plane containing a point  and with a normal vector  is 

,

Using what we were given, we get 

.

Through algebra, we get 

.

Using the formula for a plane, we have

,

where the point given is  and the normal vector is .

Plugging in the known values, you get 

.

Manipulating this equation through algebra gives you the answer 

Using the formula for a plane, we have

,

where the point given is  and the normal vector is .

Plugging in the known values, you get 

.

Manipulating this equation through algebra gives you the answer 

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 unit tangent vector, we must find the tangent vector and divide it by its magnitude.

To find the tangent vector, we take the derivative of each of the components:

The derivatives were found using the following rules:

, , 

Now, we find the magnitude of the tangent vector by taking the square root of the sum of its components:

Our final answer is 

If we are given both the normal vector to the plane and a point on the plane, we can use the formula , where  and the point on the plane is . Plugging in what we know, we get . Manipulating this equation through algebra and making the variables all on one side, we get 

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