Math Problem Statement

Find the plane equidistant from three points A(1, 2, 1), B(-1, 0, 1), and O(0, 0, 0), and find the point P on the line segment between A(4, 0, 5) and B(0, 2, 1) such that the sum of distances from P to O and AB is minimized.

Solution

The problem you're referring to from the image is written in Japanese, but I can translate and help you solve it. Let's break down the two visible problems.

Problem 4:

This problem asks to find the equation of a plane that is equidistant from three points:
A(1,2,1)A(1, 2, 1), B(1,0,1)B(-1, 0, 1), and the origin O(0,0,0)O(0, 0, 0).

Solution Strategy:

To find the plane that is equidistant from these three points, we can set up the equation of the plane as:

ax+by+cz=dax + by + cz = d

For the plane to be equidistant from the three points, the perpendicular distances from each of the three points to the plane must be equal. We can use the formula for the perpendicular distance from a point (x1,y1,z1)(x_1, y_1, z_1) to a plane ax+by+cz+d=0ax + by + cz + d = 0:

Distance=ax1+by1+cz1+da2+b2+c2\text{Distance} = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

We'll substitute the coordinates of points O(0,0,0)O(0,0,0), A(1,2,1)A(1, 2, 1), and B(1,0,1)B(-1, 0, 1) into this formula and solve for the coefficients aa, bb, and cc.

Problem 5:

This problem involves finding the coordinates of a point PP on the line segment between two points A(4,0,5)A(4, 0, 5) and B(0,2,1)B(0, 2, 1), such that the sum of distances from PP to OO and from PP to ABAB is minimized.

Solution Strategy:

  1. Find the parametric equation of line AB:
    The equation of the line segment joining two points A(x1,y1,z1)A(x_1, y_1, z_1) and B(x2,y2,z2)B(x_2, y_2, z_2) can be given as:

    P(t)=(1t)A+tB=(1t)(4,0,5)+t(0,2,1)P(t) = (1-t)A + tB = (1-t)(4, 0, 5) + t(0, 2, 1)

    This will give the parametric coordinates of the point P(t)P(t).

  2. Minimize the distance function:
    We need to minimize the total distance function, which is the sum of the distance from PP to the origin and from PP to the line AB. This can be expressed mathematically and minimized using calculus techniques (finding the derivative and setting it to zero to locate the minimum).

Would you like more details on these steps? Let me know if you'd like to go deeper into solving them.


Follow-up questions:

  1. What is the perpendicular distance formula used for finding distances between points and planes?
  2. How do you find the equation of a plane given three points in space?
  3. What is the parametric equation of a line segment between two points in 3D space?
  4. How do you use calculus to minimize a function of distances?
  5. What role does symmetry play in minimizing distances in geometry?

Tip: When dealing with 3D geometry problems, visualize the problem by sketching or mentally picturing the points, lines, and planes to better understand their spatial relationships.

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Coordinate Geometry
Planes
Vectors
Parametric Equations

Formulas

Plane equation: ax + by + cz = d
Perpendicular distance from a point to a plane: |ax1 + by1 + cz1 + d| / √(a^2 + b^2 + c^2)
Parametric equation of a line segment: P(t) = (1 - t)A + tB

Theorems

Perpendicular distance formula
Equation of a plane through three points

Suitable Grade Level

Grades 10-12