My Math Forum intersection of two lines (vectors)

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 April 7th, 2014, 10:27 AM #1 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 intersection of two lines (vectors) I have been stuck on this question for a while now, can anybody help ? Vector equation of line L1=(9,2,-1) + M(3,-4,0) Vector equation of line L2=(2,3,2) + T(a,3,b) These lines are perpendicular and intersect at point P. Find unknowns a and b. Find point of intersection P. I have not came across one of these questions with two uknowns before, only one. Can anybody give me a hint into how to solve this ? Any help much appreciated.
 April 7th, 2014, 10:35 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs What is the criterion for orthogonal (or perpendicular) vectors?
 April 7th, 2014, 10:41 AM #3 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 That their scalar product is zero ? Or did you mean something else ?
 April 7th, 2014, 10:52 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs The dot product of the two direction vectors must be zero. This will allow you to find $a$...what do you get?
 April 7th, 2014, 10:54 AM #5 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 i get a=4
 April 7th, 2014, 11:00 AM #6 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 i end up with a=4 t=1 b=-3 m=-1. point of intersection as 6,6,-1 ? Thanks from MarkFL
April 7th, 2014, 11:05 AM   #7
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From: St. Augustine, FL., U.S.A.'s oldest city

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Math Focus: Calculus/ODEs
Quote:
 Originally Posted by marks2014 i get a=4
Correct!

So now we have:

$\displaystyle \textbf{L}_1=\langle 9,2,-1 \rangle+M\langle 3,-4,0 \rangle$

$\displaystyle \textbf{L}_2=\langle 2,3,2 \rangle+T\langle 4,3,b \rangle$

Next, let's let point $P$ be $(x,y,z)$. Since the two lines intersect at that point, we may state:

$\displaystyle x=3M+9=4T+2$

$\displaystyle y=-4M+2=3T+3$

$\displaystyle z=-1=bT+2$

You now have 3 unknowns in 3 linear equations...what do you find when you solve this system?

April 7th, 2014, 11:08 AM   #8
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From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
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Math Focus: Calculus/ODEs
Quote:
 Originally Posted by marks2014 i end up with a=4 t=1 b=-3 m=-1. point of intersection as 6,6,-1 ?
Correct again! Good work.

 April 7th, 2014, 11:09 AM #9 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 you must have missed the post i just sent, (6,6,-1) ?
 April 7th, 2014, 11:10 AM #10 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 Thanks alot for your help ! really good advice ! Thanks from MarkFL

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