Find an equation of a plane containing the three points (-1, 3, -1), (2, 7, -2), (2, 8, 0) in which the coefficient of x is 9.
We can begin by finding the direction vectors of two lines that lie in the plane containing the three given points.
One direction vector can be found by subtracting the coordinates of one of the given points from another given point. Let's subtract the coordinates of (-1, 3, -1) from (2, 7, -2):
(2-(-1), 7-3, -2-(-1)) = (3, 4, -1)
Another direction vector can be found by subtracting the coordinates of a different given point from the remaining given point. Let's subtract the coordinates of (-1, 3, -1) from (2, 8, 0):
(2-(-1), 8-3, 0-(-1)) = (3, 5, 1)
Now, we can find the cross product of these two direction vectors to obtain the normal vector of the plane:
(3, 4, -1) × (3, 5, 1) = (-1, -4, 3)
Let (x, y, z) represent a generic point on the plane. We know that the direction vector (9, 0, 0) is parallel to the plane. Therefore, a normal vector of the plane will satisfy the equation:
-1(x - (-1)) - 4(y - 3) + 3(z - (-1)) = 0
Simplifying the equation, we have:
-x + 1 - 4y + 12 + 3z - 3 = 0
-x - 4y + 3z + 10 = 0
Thus, an equation of the plane containing the three given points and with a coefficient of x equal to 9 is:
9(-x - 4y + 3z + 10) = 0
-9x - 36y + 27z + 90 = 0
To find the equation of a plane containing three points, we can use the formula:
Ax + By + Cz + D = 0,
where (x, y, z) are the coordinates of a point on the plane, and A, B, C, D are constants.
First, we need to find the direction vectors of two lines in the plane, which can be calculated using two of the given points. Let's choose the points (-1, 3, -1) and (2, 7, -2).
The vector between these two points is:
<2 - (-1), 7 - 3, -2 - (-1)> = <3, 4, -1>.
Similarly, let's find the vector between the points (-1, 3, -1) and (2, 8, 0):
<2 - (-1), 8 - 3, 0 - (-1)> = <3, 5, 1>.
Now, we can find the cross product of these two vectors:
<3, 4, -1> x <3, 5, 1> = <(-4) - 5, (3)(1) - (-1)(3), (3)(5) - (4)(3)> = <-9, 6, 3>.
This cross product vector, <-9, 6, 3>, will be orthogonal to the plane.
Now, let's substitute the coordinates of one of the given points, such as (-1, 3, -1), into the equation Ax + By + Cz + D = 0:
(-9)(-1) + 6(3) + 3(-1) + D = 0,
9 + 18 - 3 + D = 0,
24 + D = 0,
D = -24.
So, the equation of the plane that contains the three points (-1, 3, -1), (2, 7, -2), (2, 8, 0), and has a coefficient of x equal to 9 is:
-9x + 6y + 3z - 24 = 0.