Determine an equation for the plane that contains the points A(6, 2, 3), B(5, 1, -3), and C(5, 7, -2).
|(x-6) (y-2) (z-3)|
|(5-6) (1-2) (-3-3)| = 0
|(5-6) (7-2) (-2-3)|
35(x-6) + (y-2) - 6(z-3) = 0
35x + y - 6z = 194
thank you :)
To determine an equation for the plane that contains the given points A(6, 2, 3), B(5, 1, -3), and C(5, 7, -2), you can use the concept of cross products.
First, find two vectors that lie in the plane using the given points:
Vector AB = B - A = (5 - 6, 1 - 2, -3 - 3) = (-1, -1, -6)
Vector AC = C - A = (5 - 6, 7 - 2, -2 - 3) = (-1, 5, -5)
Next, take the cross product of these two vectors:
Vector AB x AC = (-1, -1, -6) x (-1, 5, -5)
The cross product can be calculated as follows:
(x, y, z) = (ABy * ACz - ABz * ACy, ABz * ACx - ABx * ACz, ABx * ACy - ABy * ACx)
= (-1 * -5 - (-6) * 5, -6 * (-1) - (-1) * (-5), -1 * 5 - (-1) * (-1))
= (5 + 30, 6 - 5, -5 - 1)
= (35, 1, -4)
So, we have the vector (35, 1, -4). Now, to obtain the equation of the plane, substitute one of the given points (let's use point A) and the vector into the plane equation: Ax + By + Cz = D
Substituting (6, 2, 3) into the equation:
35(6) + 1(2) + (-4)(3) = D
210 + 2 - 12 = D
D = 200
Therefore, the equation for the plane that contains the points A(6, 2, 3), B(5, 1, -3), and C(5, 7, -2) is:
35x + y - 4z = 200
To determine an equation for the plane that contains the points A(6, 2, 3), B(5, 1, -3), and C(5, 7, -2), we can use the point-normal form equation of a plane.
The equation of a plane in point-normal form is given by:
Ax + By + Cz = D
To find the values of A, B, C, and D, we need to determine the coefficients from the given points A, B, and C.
1. Find the vectors AB and AC by subtracting the coordinates of the points:
AB = B - A = (5-6, 1-2, -3-3) = (-1, -1, -6)
AC = C - A = (5-6, 7-2, -2-3) = (-1, 5, -5)
2. Find the cross product of AB and AC to get the normal vector of the plane:
n = AB x AC
n = (-1, -1, -6) x (-1, 5, -5)
n = (-35, 1, 4)
3. The coefficients A, B, and C in the plane equation can be taken from the values of the normal vector. So we have:
A = -35, B = 1, C = 4
4. To find the value of D, substitute the coordinates of one of the given points (A, B, or C) into the plane equation. Let's use point A(6, 2, 3):
-35(6) + 1(2) + 4(3) = D
-210 + 2 + 12 = D
-196 = D
Therefore, the equation for the plane that contains the points A(6, 2, 3), B(5, 1, -3), and C(5, 7, -2) is:
-35x + y + 4z = -196