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

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:

The derivatives were found using the following rules:

, 

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 vector by its magnitude:

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:

The derivatives were found using the following rules:

, , 

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 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 them:

, , 

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 the line, we first must find the vector passing between the two points:

This was found by . 

Next, we pick one point, and using the formula

we get our line:

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

To start, let's find these vectors:

We found them by subtracting the terminal point (B, C) and the initial point (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  is the normal vector and  is any point on the plane.

Using our normal vector  and the point , we get the equation