Expert: Paul Klarreich - 7/5/2009

Question
QUESTION: I had try already to do this question & i know how to do it using my teacher method by forming two simultaneous equation first but i found out that this method takes a long time & complex for me. I'm asking here if you know any alternatives working which are simpler & short to solve this kind of question. I'm studying "A Level" mathematics on vectors topic. Two planes have equations 2x-y-3z=7 & x+2y+2z=0. Find a vector equation for their line of intersection. Thanks a lot

ANSWER: Country:  China
Category:  Advanced Math
Private:  No
 
Subject:  Vector
Question:  I had try already to do this question & i know how to do it using my teacher method by forming two simultaneous equation first but i found out that this method takes a long time & complex for me. I'm asking here if you know any alternatives working which are simpler & short to solve this kind of question. I'm studying "A Level" mathematics on vectors topic. Two planes have equations 2x-y-3z=7 & x+2y+2z=0. Find a vector equation for their line of intersection. Thanks a lot
........................................
Hi, Warm,

A vector equation of a line would be:

A (x - x0) = 0

where A is the direction vector (see below) and
x0 is a point on the line.   That means x0 is a point on both planes, so you will have to do some simultaneous equation stuff to find it.

Your equations are:

2x-y-3z=7
x+2y+2z=0

and since you have two equations in 3 variables, an infinite number of solutions, so we can just pick a value for any one of x,y,z.

Perhaps pick  z = 0?

2x -  y = 7
x + 2y = 0

4x - 2y = 14
x + 2y = 0

5x = 14,  x = 14/5

- y = 7 - 28/5

- y = 35/5 - 28/5

y = -7/5
................
OK, now we need a direction vector.  Take the two equations:

2x-y-3z=7  ==>  <2,-1,-3> normal vector.
x+2y+2z=0  ==>  <1,2,2> normal vector.

Now the cross-product of those normal vectors is what you want.


---------- FOLLOW-UP ----------

QUESTION: Do you know any other method to solve this question, other then the one you mentioned? Thanks a million

Answer
OK, it could go like this:

............. CONTINUATION..............
OK, now we need a direction vector.  Take the two equations:

2x-y-3z=7  ==>  <2,-1,-3> normal vector.
x+2y+2z=0  ==>  <1,2,2> normal vector.

|  I  j   k |
|  2 -1  -3 |
|  1  2   2 |

(-2 + 6)i + (-3-4)j + (4 + 1)k

4i - 7j + 5k

(4i - 7j + 5k) dot ((x,y,z) - (14/5, -7/5, 0)) = 0

4(x - 14/5) = 0
-7(y + 7/5) = 0
5(z) = 0
=========ALTERNATIVE SOLUTION======================

2x - y - 3z = 7
x + 2y + 2z = 0

Solve:

4x - 2y - 6z = 14
x  + 2y + 2z = 0
--------------------
5x - 4z = 14
   4z + 14
x = -------
      5

2x - y - 3t = 7
-2x -4y -4t = 0
-----------------
 -5y - 7t = 7

 -5y = 7t + 7
      7t + 7
  y = -------
        -5

OK, set a parameter, such as:
let  t = z/5, and then:

x = 4t + 14/5

y = -7t - 7/5

z = 5t,

and you have your equation:

<4,-7,5>(<x,y,z> - <14/5,-7/5,0>) = 0