Math Problem Statement

The shortest distance between the point P(1m, 4m, 3m) and Q(-2m, -1m, 2m) is:

Solution

The question asks to find the shortest distance between two points, P(1 m,4 m,3 m)P(1 \text{ m}, 4 \text{ m}, 3 \text{ m}) and Q(2 m,1 m,2 m)Q(-2 \text{ m}, -1 \text{ m}, 2 \text{ m}).

The shortest distance between two points in 3D space is given by the formula for the Euclidean distance: d=(x2x1)2+(y2y1)2+(z2z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Substitute the coordinates of points P(1,4,3)P(1, 4, 3) and Q(2,1,2)Q(-2, -1, 2) into the formula:

d=((21)2)+((14)2)+((23)2)d = \sqrt{((-2 - 1)^2) + ((-1 - 4)^2) + ((2 - 3)^2)} d=(3)2+(5)2+(1)2d = \sqrt{(-3)^2 + (-5)^2 + (-1)^2} d=9+25+1d = \sqrt{9 + 25 + 1} d=35d = \sqrt{35}

Thus, the shortest distance between the two points is 35\sqrt{35} meters.

The correct answer is option (4): 35\sqrt{35} meters.

Would you like further details or have any other questions?


Here are 5 related questions for deeper understanding:

  1. How is the Euclidean distance formula derived?
  2. What is the difference between 2D and 3D distance calculations?
  3. How can the distance formula be applied in physics to calculate displacement?
  4. What are some practical applications of finding distances in 3D space?
  5. How do you calculate distances in non-Euclidean geometries?

Tip: Always square the differences in coordinates carefully to avoid sign errors in distance calculations!

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Math Problem Analysis

Mathematical Concepts

Euclidean Geometry
3D Distance Calculation

Formulas

d = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12