Math Problem Statement

Find the parametric and symmetric equation of the line through the points (-1,4,-3) and (4,2,-1).

Solution

To find the parametric and symmetric equations of the line passing through the points (1,4,3)(-1, 4, -3) and (4,2,1)(4, 2, -1), follow these steps:

Step 1: Find the direction vector

The direction vector v\mathbf{v} is the vector from the first point to the second point. It is given by:

v=(x2x1,y2y1,z2z1)\mathbf{v} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)

Substituting the points (1,4,3)(-1, 4, -3) and (4,2,1)(4, 2, -1):

v=(4(1),24,1(3))=(5,2,2)\mathbf{v} = (4 - (-1), 2 - 4, -1 - (-3)) = (5, -2, 2)

So, the direction vector is v=(5,2,2)\mathbf{v} = (5, -2, 2).

Step 2: Parametric equations of the line

The parametric equations of a line are given by:

x=x1+tvx,y=y1+tvy,z=z1+tvzx = x_1 + t \cdot v_x, \quad y = y_1 + t \cdot v_y, \quad z = z_1 + t \cdot v_z

Here, (x1,y1,z1)(x_1, y_1, z_1) is the point (1,4,3)(-1, 4, -3), and (vx,vy,vz)(v_x, v_y, v_z) are the components of the direction vector (5,2,2)(5, -2, 2). Substituting these values:

x=1+5t,y=42t,z=3+2tx = -1 + 5t, \quad y = 4 - 2t, \quad z = -3 + 2t

Thus, the parametric equations of the line are:

x=1+5t,y=42t,z=3+2t\boxed{x = -1 + 5t, \quad y = 4 - 2t, \quad z = -3 + 2t}

Step 3: Symmetric equation of the line

To find the symmetric equation, solve each parametric equation for tt:

  1. From x=1+5tx = -1 + 5t, we get t=x+15t = \frac{x + 1}{5}.
  2. From y=42ty = 4 - 2t, we get t=4y2t = \frac{4 - y}{2}.
  3. From z=3+2tz = -3 + 2t, we get t=z+32t = \frac{z + 3}{2}.

Now, equate all these expressions for tt:

x+15=4y2=z+32\frac{x + 1}{5} = \frac{4 - y}{2} = \frac{z + 3}{2}

Thus, the symmetric equation of the line is:

x+15=4y2=z+32\boxed{\frac{x + 1}{5} = \frac{4 - y}{2} = \frac{z + 3}{2}}

Would you like more details on any step or clarification?


Here are 5 questions to expand your knowledge:

  1. How can you find the equation of a line in vector form using parametric equations?
  2. How would you determine if two lines in space are parallel?
  3. Can parametric equations represent curves other than lines?
  4. What is the physical interpretation of the direction vector in 3D space?
  5. How can you derive the Cartesian equation of a plane from three points?

Tip: When solving geometric problems in 3D, the direction vector plays a crucial role in defining lines and angles between them.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
3D Vectors
Lines in Space

Formulas

Direction vector formula: v = (x2 - x1, y2 - y1, z2 - z1)
Parametric equations of a line: x = x1 + t * vx, y = y1 + t * vy, z = z1 + t * vz
Symmetric equations of a line: (x - x1)/vx = (y - y1)/vy = (z - z1)/vz

Theorems

Vector Representation of a Line

Suitable Grade Level

Grades 10-12