Math Problem Statement
Solution
The problem involves a cube with a side length of 6 cm, and we are asked to find the distance from point to the plane .
Steps to Solve:
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Identify Points and Plane:
- The cube has vertices labeled as .
- The plane passes through points , , and .
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Determine Coordinates: Let's assign coordinates to the vertices assuming the origin at :
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Find the Equation of Plane : The plane passes through , , and . To find the equation of the plane:
- First, find two vectors in the plane:
- The normal vector to the plane is the cross product of and : \hat{i} & \hat{j} & \hat{k} \\ 6 & 0 & 6 \\ 0 & 6 & 6 \end{array} \right| = (-36, -36, 36)$$ Simplified, $$\vec{n} = (-1, -1, 1)$$.
- The equation of the plane is:
- First, find two vectors in the plane:
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Calculate Distance from to Plane : The coordinates of point are . The distance from a point to a plane is given by: Substituting , , , , and :
Thus, the distance from point to the plane is cm.
Would you like further clarification or more details on any step?
Here are five related questions to expand on this:
- How do you find the equation of a plane given three points?
- What is the geometric meaning of the normal vector to a plane?
- How do you compute the cross product of two vectors?
- What are the properties of distances from points to planes?
- How does the Pythagorean theorem apply to 3D geometry?
Tip: When working with 3D geometry, always try to visualize the problem by assigning coordinates to key points and using vector operations to simplify calculations.
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Math Problem Analysis
Mathematical Concepts
3D Geometry
Planes
Distance from Point to Plane
Formulas
Cross product of two vectors: \vec{AF} \times \vec{AH}
Equation of a plane: Ax + By + Cz + D = 0
Distance from point to plane: d = |Ax_1 + By_1 + Cz_1 + D| / sqrt(A^2 + B^2 + C^2)
Theorems
Distance from point to plane in 3D geometry
Suitable Grade Level
Grade 10-12
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