Math Problem Statement

Найти точку, симметричную точке P(1, 0, -1) относительно прямой (x + 1) / 1 = (y - 1) / 2 = z / 3.

Solution

The problem requires finding a point symmetrical to P(1,0,1)P(1, 0, -1) relative to the given line:

x+11=y12=z3.\frac{x+1}{1} = \frac{y-1}{2} = \frac{z}{3}.

Let's solve this step by step:


1. Parametric Equation of the Line

The symmetric point is relative to the given line. To describe the line parametrically:

  • Point on the line: (1,1,0)(-1, 1, 0)
  • Direction vector: (1,2,3)(1, 2, 3)

The parametric equations are: x=1+t,y=1+2t,z=3t,x = -1 + t, \quad y = 1 + 2t, \quad z = 3t, where tt is the parameter.


2. Find the Projection of PP on the Line

To find the symmetric point, we first need the projection of P(1,0,1)P(1, 0, -1) onto the line. This involves finding tt such that the vector from PP to a point on the line is perpendicular to the line's direction vector.

  • A point on the line: (1+t,1+2t,3t)(-1 + t, 1 + 2t, 3t)
  • Vector from PP to this point: PQ=(1+t1,1+2t0,3t+1)=(t2,1+2t,3t+1).\vec{PQ} = (-1+t-1, 1+2t-0, 3t+1) = (t-2, 1+2t, 3t+1). This vector is perpendicular to the direction vector (1,2,3)(1, 2, 3) if their dot product is zero: (t2)(1)+(1+2t)(2)+(3t+1)(3)=0.(t-2)(1) + (1+2t)(2) + (3t+1)(3) = 0. Simplify: t2+2+4t+9t+3=0    14t+3=0    t=314.t - 2 + 2 + 4t + 9t + 3 = 0 \implies 14t + 3 = 0 \implies t = -\frac{3}{14}.

3. Coordinates of the Projection Point

Substitute t=314t = -\frac{3}{14} into the parametric equations: x=1+t=1314=1714,x = -1 + t = -1 - \frac{3}{14} = -\frac{17}{14}, y=1+2t=1+2(314)=1614=814=47,y = 1 + 2t = 1 + 2\left(-\frac{3}{14}\right) = 1 - \frac{6}{14} = \frac{8}{14} = \frac{4}{7}, z=3t=3(314)=914.z = 3t = 3\left(-\frac{3}{14}\right) = -\frac{9}{14}. The projection point is: Q(1714,47,914).Q\left(-\frac{17}{14}, \frac{4}{7}, -\frac{9}{14}\right).


4. Symmetrical Point Calculation

The symmetrical point PP' is found using: P=2QP.P' = 2Q - P. Substitute the coordinates of QQ and P(1,0,1)P(1, 0, -1): x=2(1714)1=34141414=4814=247,x' = 2\left(-\frac{17}{14}\right) - 1 = -\frac{34}{14} - \frac{14}{14} = -\frac{48}{14} = -\frac{24}{7}, y=2(47)0=87,y' = 2\left(\frac{4}{7}\right) - 0 = \frac{8}{7}, z=2(914)(1)=1814+1414=414=27.z' = 2\left(-\frac{9}{14}\right) - (-1) = -\frac{18}{14} + \frac{14}{14} = -\frac{4}{14} = -\frac{2}{7}. The symmetrical point is: P(247,87,27).P'\left(-\frac{24}{7}, \frac{8}{7}, -\frac{2}{7}\right).


Would you like further clarification or details? 😊


Related Questions:

  1. How do you calculate the parametric equations of a line in 3D?
  2. What is the mathematical meaning of the dot product in geometry?
  3. How can you verify if a point lies on a given line in 3D space?
  4. What are alternative methods for finding a symmetric point relative to a line?
  5. How is the projection of a point onto a line useful in real-world applications?

Tip:

Always check your calculations by substituting the final point back into the geometric constraints to confirm symmetry or alignment!

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Symmetry in Space
Parametric Equations
Vector Projections

Formulas

Parametric equation of a line: x = x0 + at, y = y0 + bt, z = z0 + ct
Dot product condition for perpendicular vectors: a1 * b1 + a2 * b2 + a3 * b3 = 0
Symmetry point formula: P' = 2Q - P

Theorems

Projection of a point onto a line in 3D
Symmetry principle in vector spaces

Suitable Grade Level

Undergraduate Math or Advanced High School (Grades 11-12)