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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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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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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 Tags intersection, lines, vectors Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Shamieh Calculus 2 January 19th, 2014 04:41 AM stgeorge Linear Algebra 6 September 19th, 2012 07:02 AM mathsm111 Algebra 3 July 3rd, 2011 10:40 AM TsAmE Calculus 7 September 9th, 2010 02:22 PM ben Algebra 2 June 21st, 2008 04:45 AM

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