Calculating Magnitudes and Ratios of Position Vectors on a Line

  • Thread starter Thread starter pavadrin
  • Start date Start date
  • Tags Tags
    Line Vector
AI Thread Summary
The discussion revolves around calculating the magnitudes and ratios of position vectors for points A, B, and C on a specified line. The correct position vectors were clarified, emphasizing the need to substitute the given values of λ into the line equation. The magnitude of segment AB was confirmed to be the square root of 17, while the magnitude of segment BC was initially miscalculated as the square root of 20. The ratio AB:BC was explained as simply the ratio of the lengths of the two segments, which should yield a value of two. The original poster expressed gratitude for the clarification and indicated they could proceed with the calculations.
pavadrin
Messages
154
Reaction score
0
Points A, B and C lie on the line r = 2i + 3j + λ(i + 4j) and have position vectors given by λ=-1, λ=1 and λ=2 respectively.
Find a) AB
b) AB:BC
The answer for part a is the square root of 17 (in exact form) however I keep getting the suare root of 20.
Here is what I am doing:
OB = 3i – j and OC = 5i – 5j
BC= BO + OC
BC= -(3i – j) + (5i – 5j)
BC= 2i – 4j
Therefore the magnitude of BC = square root of 20
As for part b, I’m not sure what it’s asking me to do
Thanks
Pavadrin
 
Last edited:
Physics news on Phys.org
pavadrin said:
Points A, B and C lie on the line r = 2i + 3j + λ(i + 4j) and have position vectors given by λ=-1, λ=1 and λ=2 respectively.
Find a) AB
b) AB:BC
The answer for part a is the square root of 17 (in exact form) however I keep getting the suare root of 20.
Here is what I am doing:
OB = 3i – j and OC = 5i – 5j
BC= BO + OC
BC= -(3i – j) + (5i – 5j)
BC= 2i – 4j
Therefore the magnitude of BC = square root of 20
As for part b, I’m not sure what it’s asking me to do
Thanks
Pavadrin

I don't get it. The first part is asking for the magnitude of AB, yet you seem to be working out the magnitude of BC?:confused:

In any case, your position vectors for both OB and OC are wrong. How did you do that? All you have to do is substitute the given values of λ into the equation to get the position vectors.

Maybe I think it'd be better if you showed full working on how you get OA, OB and OC, because the first part is simple algebra.

Part b) AB:BC is simply asking you to find the ratio between the lengths of the two line segments AB and BC. All you have to do is find |AB|, |BC| then take the ratio (I get two).
 
sorry, it was meant to say the magnitude of BC, not AB. I have solved this now, so there's no need to worry bout it. As for part b), thanks for clearing that piont up, i should be able to work from here, otherwise i will re-post asking fro help
thanks
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top