Determine the Cartesian equation of each of the following planes:
b) through the points (3,0,1) and (o,1,-1), and perpendicular to the plane with equation x-y-z+1=0
My teacher told me to find a direction vector, cross that with the normal, and set up my cartesian equation. I don't really understand why we do this though?
To determine the Cartesian equation of a plane, one common approach is to find a normal vector to the plane and use it in the equation of a plane, which is usually in the form of Ax + By + Cz + D = 0.
The reason we find a direction vector and cross it with the normal vector is to ensure that the resulting vector is perpendicular (or orthogonal) to both the direction vector and the normal vector. This approach helps us specify a unique plane.
Here's the step-by-step process to determine the Cartesian equation of the given plane:
Step 1: Find the direction vector of the line passing through the two given points.
Given points: P1(3, 0, 1) and P2(0, 1, -1).
Direction vector: D = P2 - P1 = (0 - 3, 1 - 0, -1 - 1) = (-3, 1, -2)
Step 2: Find the normal vector of the given plane.
Given plane equation: x - y - z + 1 = 0
Normal vector: N = (A, B, C) = (1, -1, -1)
Step 3: Take the cross product of the direction vector and normal vector.
D × N = (-3, 1, -2) × (1, -1, -1)
= [(1 * -2) - (-1 * -2), (-3 * -1) - (1 * -1), (-3 * -1) - (1 * -1)]
= (-2 + 2, 3 + 1, 3 + 1)
= (0, 4, 4)
Step 4: Setup the Cartesian equation using the direction vector and cross product vector.
The Cartesian equation of the plane passing through the given points and perpendicular to the given plane is:
0(x - 3) + 4(y - 0) + 4(z - 1) = 0
Simplifying the equation:
4(y - 0) + 4(z - 1) = 0
4y + 4z - 4 = 0
Therefore, the Cartesian equation of the plane is 4y + 4z - 4 = 0.
To determine the Cartesian equation of a plane, you need to find a point on the plane and a normal vector to the plane. Once you have these, you can use the general equation of a plane to write its Cartesian equation.
Now let's apply this to your specific problem.
Given two points on the plane (3,0,1) and (0,1,-1), we can find a direction vector that lies on the plane by subtracting the coordinates of these two points. Let's call this vector "v":
v = (3-0, 0-1, 1-(-1))
= (3, -1, 2)
Now, since the plane is perpendicular to the plane with the equation x-y-z+1=0, the normal vector of our desired plane will be parallel to the normal vector of the given plane. The given plane's normal vector can be read directly from its equation as (1, -1, -1).
Next, we need to find the cross product of the direction vector "v" and the normal vector of the given plane. The cross product of two vectors gives us another vector that is perpendicular to both of the original vectors. Hence, this cross product will be normal to our desired plane.
n = v x (1, -1, -1)
= (3, -1, 2) x (1, -1, -1)
= (1, 1, -2)
Now we have a point (3, 0, 1) on the plane and a normal vector (1, 1, -2) to the plane. We can use the Cartesian equation of a plane, which is given by:
Ax + By + Cz + D = 0
Substituting the values, we get:
1(x - 3) + 1(y - 0) - 2(z - 1) = 0
x + y - 2z + 2 = 0
which is the Cartesian equation of the desired plane.
Hence, the Cartesian equation of the plane passing through the points (3,0,1) and (0,1,-1), and perpendicular to the plane with equation x-y-z+1=0 is x + y - 2z + 2 = 0.
see a good discussion of a similar problem here:
http://math.stackexchange.com/questions/199475/find-an-equation-of-the-plane-passing-through-2-points-and-perpendicular-to-anot
google is your friend.