Math Problem Statement

Составить общее уравнение плоскости, которая проходит через прямую \( \frac{x}{3} = \frac{y}{3} = \frac{z-1}{2} \) и перпендикулярна плоскости \( 2x + 2y + 3z = 7 \).

Solution

This task requires constructing the general equation of a plane that passes through the given line and is perpendicular to the plane 2x+2y+3z=72x + 2y + 3z = 7.

Given Information:

  1. The line is described as: x3=y3=z−12.\frac{x}{3} = \frac{y}{3} = \frac{z - 1}{2}. This provides a parametric representation: x=3t,y=3t,z=1+2t.x = 3t, \quad y = 3t, \quad z = 1 + 2t. Hence, a point on the line is (3,3,1)(3, 3, 1), and the direction vector of the line is d⃗=(3,3,2)\vec{d} = (3, 3, 2).

  2. The plane to be constructed is perpendicular to the plane 2x+2y+3z=72x + 2y + 3z = 7, so its normal vector will also be perpendicular to the vector (2,2,3)(2, 2, 3), which is the normal vector of the given plane.

Step 1: Finding a normal vector for the desired plane

The desired plane should contain the line and be perpendicular to the given plane. Therefore, its normal vector should be perpendicular to both:

  • (2,2,3)(2, 2, 3), the normal vector of the given plane.
  • (3,3,2)(3, 3, 2), the direction vector of the given line.

We calculate the cross product of these vectors: n⃗=d1⃗×d2⃗,\vec{n} = \vec{d_1} \times \vec{d_2}, where: d1⃗=(2,2,3),d2⃗=(3,3,2).\vec{d_1} = (2, 2, 3), \quad \vec{d_2} = (3, 3, 2).

The cross product is computed as:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 3 \\ 3 & 3 & 2 \end{vmatrix} = \mathbf{i} \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 2 \\ 3 & 3 \end{vmatrix}.$$ Expanding: $$\vec{n} = \mathbf{i} (2 \cdot 2 - 3 \cdot 3) - \mathbf{j} (2 \cdot 2 - 3 \cdot 3) + \mathbf{k} (2 \cdot 3 - 2 \cdot 3),$$ $$\vec{n} = \mathbf{i} (-5) - \mathbf{j} (-5) + \mathbf{k} (0).$$ Thus: $$\vec{n} = (-5, 5, 0).$$ --- ### Step 2: Equation of the plane The plane passes through the point $$(3, 3, 1)$$ and has a normal vector $$(-5, 5, 0)$$. Its equation is: $$-5(x - 3) + 5(y - 3) + 0(z - 1) = 0.$$ Simplifying: $$-5x + 15 + 5y - 15 = 0,$$ $$-5x + 5y = 0.$$ Finally: $$x - y = 0.$$ --- ### Final Answer: The equation of the desired plane is: $$x - y = 0.$$ Would you like me to explain any step further or verify this solution? Here are 5 related questions to deepen your understanding: 1. How is the direction vector of a line used to construct a plane? 2. Why is the cross product of two vectors used to find the normal vector of a plane? 3. How would the solution change if the line was parameterized differently? 4. What are the properties of perpendicular planes in three-dimensional geometry? 5. Can this method be extended to higher dimensions? **Tip:** Always verify the perpendicularity condition by checking the dot product of the normal vectors of the two planes.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Planes in 3D Space
Cross Product

Formulas

Equation of a plane: \( A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \)
Cross product formula

Theorems

Perpendicularity of planes using dot product

Suitable Grade Level

Grades 11-12

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