Math Problem Statement

obtain the Cartesian equation of the plane that passes through the points (3, 5, 1), (2, 2, 2) and (−2, −2, 0).

Solution

To find the Cartesian equation of the plane passing through the points (3,5,1)(3, 5, 1), (2,2,2)(2, 2, 2), and (2,2,0)(-2, -2, 0), follow these steps:


Step 1: Find two direction vectors in the plane

Given points, we form two vectors that lie in the plane: v1=(23,25,21)=(1,3,1)\vec{v_1} = (2 - 3, 2 - 5, 2 - 1) = (-1, -3, 1) v2=(23,25,01)=(5,7,1)\vec{v_2} = (-2 - 3, -2 - 5, 0 - 1) = (-5, -7, -1)


Step 2: Compute the normal vector

The normal vector to the plane, n\vec{n}, is perpendicular to both v1\vec{v_1} and v2\vec{v_2}. Compute it using the cross product: n=v1×v2\vec{n} = \vec{v_1} \times \vec{v_2}

The determinant for the cross product is:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -3 & 1 \\ -5 & -7 & -1 \end{vmatrix}$$ Expanding: $$\vec{n} = \mathbf{i} \begin{vmatrix} -3 & 1 \\ -7 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -1 & 1 \\ -5 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -1 & -3 \\ -5 & -7 \end{vmatrix}$$ Simplify the minors: $$\vec{n} = \mathbf{i}((-3)(-1) - (1)(-7)) - \mathbf{j}((-1)(-1) - (-5)(1)) + \mathbf{k}((-1)(-7) - (-3)(-5))$$ $$\vec{n} = \mathbf{i}(3 + 7) - \mathbf{j}(1 + 5) + \mathbf{k}(-7 - 15)$$ $$\vec{n} = 10\mathbf{i} - 6\mathbf{j} - 22\mathbf{k}$$ Thus, the normal vector is: $$\vec{n} = (10, -6, -22)$$ --- ### **Step 3: Form the Cartesian equation of the plane** The general equation of a plane is: $$10x - 6y - 22z = d$$ Substitute the coordinates of a point (e.g., $$(3, 5, 1)$$) into the equation to find $$d$$: $$10(3) - 6(5) - 22(1) = d$$ $$30 - 30 - 22 = d$$ $$d = -22$$ Thus, the equation of the plane is: $$10x - 6y - 22z = -22$$ --- ### **Final Answer** The Cartesian equation of the plane is: $$10x - 6y - 22z + 22 = 0$$ Would you like further explanation or assistance with any step? 😊 --- ### **Related Questions** 1. How do you compute the cross product of two vectors geometrically? 2. What does the normal vector of a plane represent in 3D space? 3. How can the distance from a point to a plane be calculated? 4. What are the parametric equations of the same plane? 5. How can you verify if a given point lies on a plane? --- ### **Tip** When computing the cross product, carefully organize the determinant's expansion to avoid sign errors.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cartesian Geometry
Planes in 3D Space

Formulas

Direction vector formula: v = (x2 - x1, y2 - y1, z2 - z1)
Cross product formula for normal vector: n = v1 × v2
Cartesian equation of a plane: ax + by + cz = d

Theorems

Cross Product Theorem
Plane Equation Theorem

Suitable Grade Level

Grades 10-12