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April 7th, 2014, 10:27 AM   #1
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Question 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.
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April 7th, 2014, 10:35 AM   #2
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What is the criterion for orthogonal (or perpendicular) vectors?
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April 7th, 2014, 10:41 AM   #3
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That their scalar product is zero ? Or did you mean something else ?
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April 7th, 2014, 10:52 AM   #4
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The dot product of the two direction vectors must be zero. This will allow you to find $a$...what do you get?
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April 7th, 2014, 10:54 AM   #5
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i get a=4
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April 7th, 2014, 11:00 AM   #6
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i end up with a=4 t=1 b=-3 m=-1.

point of intersection as 6,6,-1 ?
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April 7th, 2014, 11:05 AM   #7
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Quote:
Originally Posted by marks2014 View Post
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?
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April 7th, 2014, 11:08 AM   #8
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i end up with a=4 t=1 b=-3 m=-1.

point of intersection as 6,6,-1 ?
Correct again! Good work.
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April 7th, 2014, 11:09 AM   #9
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you must have missed the post i just sent, (6,6,-1) ?
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April 7th, 2014, 11:10 AM   #10
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Thanks alot for your help ! really good advice !
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