Math Problem Statement

  1. Write parametric equations of each given line: a) the line in R 3 passing through (11, −6, 7) parallel to vector 3i + j − 12k b) the line in R 3 passing through the points (3, 1, 3) and (2, −4, −2) c) the x-axis (in R 2 )
  2. Write parametric equations of each given line: a) the line in R 4 passing through the points (−1, 0, 3, −12) and (6, −15, 11, −8) b) the line in R 2 whose slope-intercept equation is y = −1 3 x − 5 3 c) the x-axis (in R 3 )

Solution

Let's tackle the problem by writing parametric equations for each of the lines as described.

55. Parametric Equations in R3\mathbb{R}^3:

(a) Line passing through (11,6,7)(11, -6, 7) and parallel to the vector 3i+j12k3\mathbf{i} + \mathbf{j} - 12\mathbf{k}:

The parametric equation of a line in R3\mathbb{R}^3 is given by:

r(t)=r0+tv\mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{v}

Where:

  • r0=(x0,y0,z0)\mathbf{r}_0 = (x_0, y_0, z_0) is a point on the line, and
  • v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3) is the direction vector of the line.

Here, r0=(11,6,7)\mathbf{r}_0 = (11, -6, 7) and v=(3,1,12)\mathbf{v} = (3, 1, -12). Thus, the parametric equations of the line are:

x=11+3tx = 11 + 3t y=6+ty = -6 + t z=712tz = 7 - 12t

(b) Line passing through the points (3,1,3)(3, 1, 3) and (2,4,2)(2, -4, -2):

To find the parametric equation of a line passing through two points, we first need to determine the direction vector:

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

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

v=(23,41,23)=(1,5,5)\mathbf{v} = (2 - 3, -4 - 1, -2 - 3) = (-1, -5, -5)

Now, the parametric equations of the line through the point (3,1,3)(3, 1, 3) are:

x=3tx = 3 - t y=15ty = 1 - 5t z=35tz = 3 - 5t

(c) Line along the x-axis in R2\mathbb{R}^2:

For a line along the x-axis in R2\mathbb{R}^2, the yy-coordinate is always 0, and xx varies freely. The parametric equations are:

x=tx = t y=0y = 0


56. Parametric Equations in Rn\mathbb{R}^n:

(a) Line passing through the points (1,0,3,12)(-1, 0, 3, -12) and (6,15,11,8)(6, -15, 11, -8) in R4\mathbb{R}^4:

The direction vector v\mathbf{v} is:

v=(6(1),150,113,8(12))=(7,15,8,4)\mathbf{v} = (6 - (-1), -15 - 0, 11 - 3, -8 - (-12)) = (7, -15, 8, 4)

Now, the parametric equations through (1,0,3,12)(-1, 0, 3, -12) are:

x=1+7tx = -1 + 7t y=015ty = 0 - 15t z=3+8tz = 3 + 8t w=12+4tw = -12 + 4t

(b) Line whose slope-intercept form is y=13x53y = -\frac{1}{3}x - \frac{5}{3} in R2\mathbb{R}^2:

Here, the slope m=13m = -\frac{1}{3} and the y-intercept is 53-\frac{5}{3}. The parametric equations are:

x=tx = t y=13t53y = -\frac{1}{3}t - \frac{5}{3}

(c) Line along the x-axis in R3\mathbb{R}^3:

For a line along the x-axis in R3\mathbb{R}^3, the yy- and zz-coordinates are always 0, and xx varies freely. The parametric equations are:

x=tx = t y=0y = 0 z=0z = 0


Would you like any details or have questions on this? Here are 5 related questions:

  1. How do we determine the direction vector between two points?
  2. What is the difference between parametric and symmetric equations of a line?
  3. Can a line in Rn\mathbb{R}^n be represented by more than one set of parametric equations?
  4. How does the parametric equation of a line differ when the slope is given?
  5. How do parametric equations generalize for higher dimensions like R4\mathbb{R}^4?

Tip: In parametric equations, the parameter tt typically represents how far along the line you move, making it a powerful tool for visualizing and working with lines in multiple dimensions.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Vector Geometry
Coordinate Systems

Formulas

r(t) = r0 + t * v
Direction vector v = (x2 - x1, y2 - y1, z2 - z1)

Theorems

Parametric equations of a line in vector form

Suitable Grade Level

Undergraduate Mathematics or Advanced High School