determine an equation for each plane
a) contains the point A(6,2,3), B(5,1,-3), and C(5,7,-2)
b) parallel to 2x + 6y + 4z = 1 and contains the point P(3,2,1)
a) To determine an equation for a plane, we need to find the normal vector of the plane and any point on the plane.
First, we can find two vectors on the plane using points A, B, and C.
Let's calculate two vectors on the plane:
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, we can find the normal vector of the plane by taking the cross product of vectors AB and AC.
Normal vector = AB x AC
= (-1, -1, -6) x (-1, 5, -5)
To calculate the cross product, we can use the determinant method:
i j k
-1 -1 -6
-1 5 -5
= ( (-1 * -5) - (-6 * 5), (-1 * -5) - (-1 * -5), (-1 * -5) - (-1 * -5) )
= (25 - (-30), 5 - (-5), 5 - (-5))
= (25 + 30, 5 + 5, 5 + 5)
= (55, 10, 10)
We now have the normal vector of the plane, which is (55, 10, 10).
To find the equation of the plane, we can substitute the values of the normal vector and the coordinates of point A(6,2,3) into the general form of the equation of a plane:
Ax + By + Cz = D
Plugging in the values, we have:
55x + 10y + 10z = D
Now we can find the value of D by substituting the coordinates of point A into the equation:
55(6) + 10(2) + 10(3) = D
330 + 20 + 30 = D
D = 380
The equation of the plane containing points A, B, and C is:
55x + 10y + 10z = 380
b) To determine an equation for a plane parallel to a given plane, we can use the same normal vector of the given plane.
The equation 2x + 6y + 4z = 1 is in the form Ax + By + Cz = D, where A = 2, B = 6, C = 4, and D = 1.
Since the plane we want is parallel, it will have the same normal vector. So, the normal vector of the given plane is (2, 6, 4).
Next, we need a point on the plane. The point P(3,2,1) is given as a point on the plane.
Using the normal vector (2, 6, 4) and the point P(3,2,1), we can write the equation of the plane in the form Ax + By + Cz = D:
2(x - 3) + 6(y - 2) + 4(z - 1) = 0
Simplifying the equation gives:
2x - 6 + 6y - 12 + 4z - 4 = 0
2x + 6y + 4z - 22 = 0
So, the equation of the plane parallel to 2x + 6y + 4z = 1 and containing the point P(3,2,1) is:
2x + 6y + 4z - 22 = 0