Math Problem Statement

Find the equation of the plane passing through the points P(-4, -1, -2), Q(5, 2, 8), and R(-1, 0, 5). Write your equation in the form ax + by + cz + d = 0.

Solution

To find the equation of a plane passing through three points, P(4,1,2)P(-4, -1, -2), Q(5,2,8)Q(5, 2, 8), and R(1,0,5)R(-1, 0, 5), we can proceed with the following steps:

  1. Calculate Two Direction Vectors:

    • Vector PQ\vec{PQ} from PP to QQ: PQ=QP=(5(4),2(1),8(2))=(9,3,10)\vec{PQ} = Q - P = (5 - (-4), 2 - (-1), 8 - (-2)) = (9, 3, 10)
    • Vector PR\vec{PR} from PP to RR: PR=RP=(1(4),0(1),5(2))=(3,1,7)\vec{PR} = R - P = (-1 - (-4), 0 - (-1), 5 - (-2)) = (3, 1, 7)
  2. Find the Cross Product PQ×PR\vec{PQ} \times \vec{PR}: The cross product of PQ\vec{PQ} and PR\vec{PR} will give us a vector that is perpendicular to the plane, which we can use as the normal vector n=(a,b,c)\vec{n} = (a, b, c).

    PQ×PR=i^j^k^9310317\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & 3 & 10 \\ 3 & 1 & 7 \end{vmatrix} Expanding this determinant: PQ×PR=i^(37101)j^(97103)+k^(9133)\vec{PQ} \times \vec{PR} = \hat{i}(3 \cdot 7 - 10 \cdot 1) - \hat{j}(9 \cdot 7 - 10 \cdot 3) + \hat{k}(9 \cdot 1 - 3 \cdot 3) =i^(2110)j^(6330)+k^(99)= \hat{i}(21 - 10) - \hat{j}(63 - 30) + \hat{k}(9 - 9) =i^(11)j^(33)+k^(0)= \hat{i}(11) - \hat{j}(33) + \hat{k}(0) =(11,33,0)= (11, -33, 0)

    So, the normal vector n=(11,33,0)\vec{n} = (11, -33, 0).

  3. Form the Equation of the Plane: Using the point-normal form of the plane, a(xx1)+b(yy1)+c(zz1)=0a(x - x_1) + b(y - y_1) + c(z - z_1) = 0, where (x1,y1,z1)(x_1, y_1, z_1) is a point on the plane (let’s use P(4,1,2)P(-4, -1, -2)) and n=(11,33,0)\vec{n} = (11, -33, 0):

    11(x+4)33(y+1)+0(z+2)=011(x + 4) - 33(y + 1) + 0(z + 2) = 0 Expanding this: 11x+4433y33=011x + 44 - 33y - 33 = 0 11x33y+11=011x - 33y + 11 = 0

  4. Simplify the Equation: Divide through by 11 to simplify: x3y+1=0x - 3y + 1 = 0

Thus, the equation of the plane passing through points PP, QQ, and RR is: x3y+1=0x - 3y + 1 = 0

Would you like a detailed explanation of each step or have any questions about the process?

Here are five questions to expand your understanding:

  1. How can the cross product of two vectors help in determining the equation of a plane?
  2. What would change if we used a different point on the plane in the point-normal form?
  3. Why is it necessary to simplify the equation after forming it initially?
  4. How would you confirm if a fourth point lies on the same plane?
  5. Can this approach be extended to higher-dimensional spaces?

Tip: Always ensure to check the result by plugging back at least one point into the final equation to verify its correctness.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Plane Equation
Cross Product

Formulas

Vector PQ = Q - P
Vector PR = R - P
Cross Product formula for vectors
Equation of plane in point-normal form: a(x - x1) + b(y - y1) + c(z - z1) = 0

Theorems

Properties of cross products
Point-normal form of a plane equation

Suitable Grade Level

Grades 11-12