Find a vector perpendicular to the plane containing the given points.
(0,-2,2), (1,2,-3), and (4,0,-1)
My teacher was unable to finish the lesson, but I'd like to know how to solve this problem.
Use pairs of points to find two different direction vectors in the plane.
eg. (1,4,-5) and (4,2,-3)
Now take the cross product of those two vectors.
You should have learned an algorithm to do that.
My perpendicular vector in simplest form was
[2,17,14]
To find a vector perpendicular to the plane containing the given points, you can use the cross product of two vectors in the plane.
Step 1: Find two vectors in the plane. Let's choose vector A and B.
Vector A can be formed by subtracting the coordinates of the second point from the first point:
A = (1, 2, -3) - (0, -2, 2) = (1, 4, -5)
Vector B can be formed by subtracting the coordinates of the third point from the first point:
B = (4, 0, -1) - (0, -2, 2) = (4, 2, -3)
Step 2: Calculate the cross product of vectors A and B.
The cross product of two vectors can be found by performing the following calculation:
N = A x B = (A_y * B_z - A_z * B_y, A_z * B_x - A_x * B_z, A_x * B_y - A_y * B_x)
Applying the formula, we get:
N = (4 * -3 - (-5) * 2, -5 * 4 - 1 * (-3), 1 * 2 - 4 * 4)
= (-12 + 10, -20 + 3, 2 - 16)
= (-2, -17, -14)
Step 3: Normalize the vector.
To obtain a unit vector perpendicular to the plane, we need to normalize the vector N by dividing each component by the length of N.
The length of N is given by the formula: |N| = sqrt(N_x^2 + N_y^2 + N_z^2)
Applying the formula, we get:
|N| = sqrt((-2)^2 + (-17)^2 + (-14)^2)
= sqrt(4 + 289 + 196)
= sqrt(489)
≈ 22.12
Now, the unit vector U in the direction of N is obtained by dividing each component of N by its length:
U = N / |N| ≈ (-2/22.12, -17/22.12, -14/22.12)
≈ (-0.090, -0.769, -0.632)
So, a vector perpendicular to the plane containing the given points is approximately (-0.090, -0.769, -0.632).