My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
July 9th, 2016, 05:17 AM   #1
Newbie
 
Joined: Jan 2016
From: Naples

Posts: 4
Thanks: 0

Exercise on two straight lines

I have this exercise that I really do not understand how to do.
In a Euclidean space $E^3$, we have an orthonormal system and considering two straight lines $r$ and $s$ of the following equations:
$r:\{(2y + 5z - 17 = 0),(x + 1 = 0)\}$
$s:\{(4x + (a - 4)z + 16 - 3a = 0),(4y + (a + 6)z - 7a - 18 = 0)\}$
with $a$ real parameter.
1) Determine the reciprocal position of $r$ and $s$ when $a$ varies.
2) For what values of $a$ are the two straight lines orthogonal?
3) In case $r$ and $s$ are coplanar, determine the plane that contains both.

I know that in order to find the reciprocal position I need the two directional vectors of the straight lines, find a point that for each straight line, and then calculate the vector product of the two directional vectors. But in this case I have a parameter that can vary. How should I proceed in this case?

Last edited by skipjack; July 23rd, 2016 at 08:30 AM.
Kernul is offline  
 
July 13th, 2016, 04:15 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Just leave the parameter in the calculation. The first line has x = -1 and, taking parameter t= 10z, 2y + 5z - 17 = 0 becomes 2y + 2t - 17 = 0 so y = 17 - t. Parametric equations are x = -1, y = 17- t, z = t/10. The "direction vector" of the first line is <0, -1, 1/10>.

For the second line we have 4x = (4 - a)z + 3a - 16 so x = (1 - a/4)z + (3/4)a - 4 and 4y = -(a + 6)z + 7a + 18 so y = -(a/4 + 3/2)z + 7a/4 + 9/2 so parametric equations are x = (1 - a/4)t - 4, y = -(a/4 + 3/2)t + 7a/4 + 9/2, z = t. Its "direction vector" is <1- a/4, -(a/4+ 3/2), 1>.

I frankly don't know what "reciprocal position" of two lines means and your description doesn't make sense. The two lines will be orthogonal when the dot product of their directional vectors is 0. Take the dot product of the two vectors, set it equal to 0 and solve for a. To find the equation of the plane containing the two lines, take the cross product of the two direction vectors. That will give the normal vector to the plane. The equation of a plane containing point $\displaystyle (x_0, y_0, z_0)$ and with normal vector <A, B, C> is $\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.

Last edited by skipjack; July 23rd, 2016 at 08:34 AM.
Country Boy is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
exercise, lines, straight



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
straight lines yash0270 Algebra 5 September 4th, 2015 06:35 AM
Find the sum (in straight lines)???? Opposite Geometry 1 November 14th, 2014 01:04 AM
Please for Straight lines alloy Algebra 1 November 3rd, 2012 02:27 AM
About Straight Lines Plz alloy Algebra 1 October 30th, 2012 08:56 AM
Two straight lines mikeportnoy Algebra 13 December 22nd, 2009 12:45 PM





Copyright © 2019 My Math Forum. All rights reserved.