Determine the vector and parametric equation of a plane that contains the points P( 3, -1, -2), Q(2,2,0), and R(-5,2,1) and determine if point C(4,1,-1) is on the plane.
To determine the vector and parametric equation of the plane containing the points P(3, -1, -2), Q(2, 2, 0), and R(-5, 2, 1), we can use the cross product between two vectors on the plane.
Step 1: Find two vectors on the plane:
- Vector PQ is given by subtracting the coordinates of point P from point Q: PQ = Q - P = (2 - 3, 2 - (-1), 0 - (-2)) = (-1, 3, 2).
- Vector PR is given by subtracting the coordinates of point P from point R: PR = R - P = (-5 - 3, 2 - (-1), 1 - (-2)) = (-8, 3, 3).
Step 2: Find the cross product of the vectors PQ and PR:
- Taking the cross product of PQ and PR gives the normal vector to the plane, which also indicates the direction perpendicular to the plane.
- The cross product is obtained by calculating the determinant of a 3x3 matrix as follows:
| i j k |
| -1 3 2 |
| -8 3 3 |
- Evaluating the determinant gives the components of the normal vector:
i((3 * 3) - (3 * 2)) - j((-1 * 3) - (-8 * 2)) + k((-1 * 3) - (-8 * 3))
= 3 - 16 - 9 = -22.
Therefore, the normal vector to the plane is (-22, 22, -22). This vector is directed away from the plane.
Step 3: Write the parametric equation of the plane:
- Given a point P(3, -1, -2) on the plane and the normal vector n(-22, 22, -22), the parametric equation of the plane is given by:
(x, y, z) = P + s * n,
where (x, y, z) represents any point on the plane, P is a known point on the plane, n is the normal vector, and s is a scalar.
Therefore, the parametric equation of the plane is:
(x, y, z) = (3, -1, -2) + s * (-22, 22, -22).
Step 4: Check if point C(4, 1, -1) is on the plane:
- To determine if point C is on the plane, substitute its coordinates into the parametric equation and check if the equation is satisfied for any value of s.
- Plugging in the values of C into the equation, we get:
(4, 1, -1) = (3, -1, -2) + s * (-22, 22, -22).
This equation needs to have a solution for s.
Solving this equation results in:
(4, 1, -1) = (3 - 22s, -1 + 22s, -2 - 22s).
Comparing the components of both sides, we get the following equations:
4 = 3 - 22s -> 22s = -1, so s = -1/22,
1 = -1 + 22s -> 22s = 2, so s = 2/22 = 1/11,
-1 = -2 - 22s -> 22s = -1, so s = -1/22.
Since the values of s are the same for all three equations, point C(4, 1, -1) lies on the plane.
In conclusion, the vector equation of the plane is:
(-22, 22, -22) · [x - (3, -1, -2)] = 0, where · represents the dot product.
And the parametric equation of the plane is:
(x, y, z) = (3, -1, -2) + s * (-22, 22, -22).
Moreover, point C(4, 1, -1) is on the plane.