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Parametric vector equation

  • Thread starter Jbreezy
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  • #1
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Homework Statement



The position vectors of two points A,B of a line are a = < 2 ,1 ,7> : b = < 1 , 4 , -1 >
Find the parametric vector equation for any point on the line AB using λ as the parameter.


Homework Equations



In general x = a + λb
Where b is the unit vector of b.


The Attempt at a Solution


b = b/ |b|
So b = <1/ 3√2,4/ 3√2,-1/ 3√2>
so we have
x = < 2 , 1, 7> + λ<1/ 3√2,4/ 3√2,-1/ 3√2>


This correct? Please do not give me the answer but I prefer a hint. Thank you .
 

Answers and Replies

  • #2
33,084
4,788

Homework Statement



The position vectors of two points A,B of a line are a = < 2 ,1 ,7> : b = < 1 , 4 , -1 >
Find the parametric vector equation for any point on the line AB using λ as the parameter.


Homework Equations



In general x = a + λb
Where b is the unit vector of b.
The vector b in this equation is not the same as the position vector b above. You need a vector with the same direction as your line.

The Attempt at a Solution


b = b/ |b|
So b = <1/ 3√2,4/ 3√2,-1/ 3√2>
so we have
x = < 2 , 1, 7> + λ<1/ 3√2,4/ 3√2,-1/ 3√2>


This correct? Please do not give me the answer but I prefer a hint. Thank you .
 
  • #3
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The vector b in this equation is not the same as the position vector b above. You need a vector with the same direction as your line.

I don't follow you dude. So b is not the same because I made b into a unit vector. So it is lambda times the unit vector b
I don't quite follow you. Thanks :)
 
  • #4
33,084
4,788
You have two points on your line: A( 2 ,1 ,7) and B( 1 , 4 , -1). As a vector, <1, 4, -1> is NOT in the same direction as the line from A to B. You need to find that vector; i.e. AB.
 
  • #5
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Mark44,

OK, So a = < 2 , 1 ,7 > and b = < 1 , 4 , -1 >

AB = a - b = < 2 , 1 ,7 > - < 1 , 4 , -1 > = < 1 , - 3, 2 >
Turn AB into a unit vector. So you have < 1 / sqrt(14), -3/ sqrt(14) , 2/ sqrt(14) >
Then any point x is represented by x = < 2, 1, 7 > + (lambda) < 1 / sqrt(14), -3/ sqrt(14) , 2/ sqrt(14) >

Good?Thanks
 
  • #6
33,084
4,788
Yes, that's right.
 
  • #7
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Thanks Mark44.
 
  • #8
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So if I wanted this in Cartesian coordinates then x , y , z = 2 , 1 , 7 + λ ( 1/√14, - 3/√14, 2/ √14)
I'm sorry pretend they are written as column's
Then we have
x = 2 + λ(1/√14)
y = 1 + λ(- 3/√14)
z = 7 + λ(2/ √14)

Eliminate lambda and I got.

√14(x - 2) = -(y -1)(√14)/ 3 = ( z - 7)(√14)/2
Correct?
 
  • #9
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Looks fine.
 
  • #10
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Kind of ugly equation don't you think? Thanks
 
  • #11
LCKurtz
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Mark44,

OK, So a = < 2 , 1 ,7 > and b = < 1 , 4 , -1 >

AB = a - b = < 2 , 1 ,7 > - < 1 , 4 , -1 > = < 1 , - 3, 2 >
Actually, the vector from a to b is b-a, not a-b, if that matters.
 

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