Math Problem Statement

Determine whether the lines L _ ( 1 ) and L _ ( 2 ) are parallel, skew, or intersecting. L _ ( 1 ) : ( x

3 ) / ( 1 )

( y

1 ) / (

2 )

( z

1 2 ) / (

3 ) L _ ( 2 ) : ( x

4 ) / ( 1 )

( y + 6 ) / ( 3 )

( z

1 3 ) / (

7 ) parallel skew intersecting If they intersect, find the point of intersection. ( If an answer does not exist, enter DNE. )

( x , y , z )

( , x )

Solution

We are given two lines L1L_1 and L2L_2 in parametric form. Our goal is to determine whether they are parallel, skew, or intersecting, and if they intersect, find the point of intersection.

Step 1: Express the lines in parametric form

The equations for the lines are given as:

L1:x31=y12=z123L_1: \frac{x - 3}{1} = \frac{y - 1}{-2} = \frac{z - 12}{-3} L2:x41=y+63=z137L_2: \frac{x - 4}{1} = \frac{y + 6}{3} = \frac{z - 13}{-7}

We can express each line in parametric form. Let t1t_1 be the parameter for L1L_1 and t2t_2 for L2L_2.

For L1L_1: x=3+t1x = 3 + t_1 y=12t1y = 1 - 2t_1 z=123t1z = 12 - 3t_1

For L2L_2: x=4+t2x = 4 + t_2 y=6+3t2y = -6 + 3t_2 z=137t2z = 13 - 7t_2

Step 2: Check if the lines are parallel

To check if the lines are parallel, we need to compare their direction vectors.

  • The direction vector of L1L_1 is 1,2,3\langle 1, -2, -3 \rangle.
  • The direction vector of L2L_2 is 1,3,7\langle 1, 3, -7 \rangle.

For the lines to be parallel, their direction vectors must be scalar multiples of each other. Let's compare the ratios of the corresponding components:

11,23,37\frac{1}{1}, \quad \frac{-2}{3}, \quad \frac{-3}{-7}

Since these ratios are not equal, the lines are not parallel.

Step 3: Check if the lines intersect

To check if the lines intersect, we need to find values of t1t_1 and t2t_2 such that the parametric equations for L1L_1 and L2L_2 yield the same point (x,y,z)(x, y, z).

From the xx-coordinates: 3+t1=4+t2t1=t2+13 + t_1 = 4 + t_2 \quad \Rightarrow \quad t_1 = t_2 + 1

From the yy-coordinates: 12t1=6+3t21 - 2t_1 = -6 + 3t_2 Substitute t1=t2+1t_1 = t_2 + 1 into this equation: 12(t2+1)=6+3t21 - 2(t_2 + 1) = -6 + 3t_2 12t22=6+3t21 - 2t_2 - 2 = -6 + 3t_2 12t2=6+3t2-1 - 2t_2 = -6 + 3t_2 5=5t2t2=15 = 5t_2 \quad \Rightarrow \quad t_2 = 1

Now that we have t2=1t_2 = 1, substitute this into t1=t2+1t_1 = t_2 + 1: t1=2t_1 = 2

Step 4: Verify the intersection point

Substitute t1=2t_1 = 2 into the parametric equations for L1L_1: x=3+2=5,y=12(2)=3,z=123(2)=6x = 3 + 2 = 5, \quad y = 1 - 2(2) = -3, \quad z = 12 - 3(2) = 6

Substitute t2=1t_2 = 1 into the parametric equations for L2L_2: x=4+1=5,y=6+3(1)=3,z=137(1)=6x = 4 + 1 = 5, \quad y = -6 + 3(1) = -3, \quad z = 13 - 7(1) = 6

Since both lines give the same point (5,3,6)(5, -3, 6), the lines intersect at the point (5,3,6)(5, -3, 6).

Final Answer:

The lines intersect at the point (5,3,6)(5, -3, 6).


Do you want more details or have any questions?

Here are 5 related questions to expand on this topic:

  1. How do you find the parametric equations of a line given two points?
  2. What is the geometric interpretation of skew lines?
  3. How can you determine the angle between two intersecting lines in space?
  4. What conditions must be satisfied for two lines in 3D to be parallel?
  5. How do you find the shortest distance between two skew lines?

Tip: Always double-check the direction vectors when comparing two lines for parallelism.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Parametric Equations
Line Intersections in 3D

Formulas

Parametric form of a line: x = x_0 + t*v_x, y = y_0 + t*v_y, z = z_0 + t*v_z
Checking for intersection: Equate parametric forms of both lines and solve for t

Theorems

Conditions for parallelism: Direction vectors are scalar multiples

Suitable Grade Level

Grades 10-12 (Advanced Geometry/Algebra)