Hey there! Sign in to join this conversationNew here? Join for free

C4 Vector Question about the direction of a vector Watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    Name:  Screenshot_134.png
Views: 4
Size:  178.0 KB

    Here's the solution to part c:
    Attachment 703164703166

    I get that for r=a+tb, you use the vector for M (5,1,5) since that is the point. Then you need b to be the directional vector.

    I get that they can use l as the directional vector because OL is parallel to LM. However, OL is going vertically diagonally whereas MN is going vertically downwards, so doesn't the position vector l need to be made negative to represent MN? (because although OL is parallel to LM, they're in opposite directions). If it doesn't need to be made negative and the mark scheme is correct, why don't you need to make it negative?

    Surely if OL was the directional vector to go upwards from O to L, then using the same directional vector from M would make you go upwards from M rather than downwards towards N? Can you explain please?

    Thanks a lot!
    Attached Images
     
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by vector12)
    I get that for r=a+tb, you use the vector for M (5,1,5) since that is the point. Then you need b to be the directional vector.

    I get that they can use l as the directional vector because OL is parallel to LM. However, OL is going vertically diagonally whereas MN is going vertically downwards, so doesn't the position vector l need to be made negative to represent MN? (because although OL is parallel to LM, they're in opposite directions). If it doesn't need to be made negative and the mark scheme is correct, why don't you need to make it negative?

    Surely if OL was the directional vector to go upwards from O to L, then using the same directional vector from M would make you go upwards from M rather than downwards towards N? Can you explain please?

    Thanks a lot!
    r = p + tq

    In the vector equation, t is a scalar that could be positive or negative. So as long as q is parallel to the line, you can use it as your direction vector. You could also use any other vector parallel to the line e.g.

    r = p + t(-2q)

    t can be any scalar so this also works.

    I think this is what you are talking about but let us know if you're still unsure.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Notnek)
    r = p + tq

    In the vector equation, t is a scalar that could be positive or negative. So as long as q is parallel to the line, you can use it as your direction vector. You could also use any other vector parallel to the line e.g.

    r = p + t(-2q)

    t can be any scalar so this also works.

    I think this is what you are talking about but let us know if you're still unsure.
    That's kind of what I was asking about. For the vector OL, which is the directional vector l, (2,-2,3), the direction of that vector is going from the origin towards L, which is going upwards (let's say in the positive Y direction). However, MN is parallel to OL, so you can use that directional vector.

    But the directional vector for MN should be going in the negative Y direction to get from M to N, whereas they can still use the directional vector l, or OL, which is going in the positive Y direction instead. How come that is? Using that same directional vector would suggest that you go in the vertical Y direction upwards from M, would it not, rather than going downwards?
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by vector12)
    That's kind of what I was asking about. For the vector OL, which is the directional vector l, (2,-2,3), the direction of that vector is going from the origin towards L, which is going upwards (let's say in the positive Y direction). However, MN is parallel to OL, so you can use that directional vector.

    But the directional vector for MN should be going in the negative Y direction to get from M to N, whereas they can still use the directional vector l, or OL, which is going in the positive Y direction instead. How come that is? Using that same directional vector would suggest that you go in the vertical Y direction upwards from M, would it not, rather than going downwards?
    Maybe I'm misunderstanding your question but you seem confused about what the vector equation of a line is and where it coms from.

    In this question you're finding the vector equation of the line MN. This is a line not a vector so you could also call this the line NM - it doesn't matter. As I said above, in the vector equation of a line, the direction vector is any vector parallel to the line, it can be in any direction.

    It may be worth watching this video to understand what the vector equation of a line means and how it is formed.

    If you understand all this and agree with my post but are still stuck then please try clarifying your question again.
    • Community Assistant
    • Welcome Squad
    Online

    20
    ReputationRep:
    (Original post by vector12)
    That's kind of what I was asking about. For the vector OL, which is the directional vector l, (2,-2,3), the direction of that vector is going from the origin towards L, which is going upwards (let's say in the positive Y direction). However, MN is parallel to OL, so you can use that directional vector.

    But the directional vector for MN should be going in the negative Y direction to get from M to N, whereas they can still use the directional vector l, or OL, which is going in the positive Y direction instead. How come that is? Using that same directional vector would suggest that you go in the vertical Y direction upwards from M, would it not, rather than going downwards?
    Do you agree that \overrightarrow{MN}=-(\overrightarrow{OL}) ?

    So your eq. for r is just r=(5,1,5)+t(\overrightarrow{MN}) as you'd say, correct?

    But note that this is just the same as r=(5,1,5)+t(-\overrightarrow{OL})=(5,1,5)+(-t)(\overrightarrow{OL})

    Since t can be any number in \mathbb{R}, we can just get rid off the negative (as it doesn't really matter) by saying that (-t) \mapsto T, and say that:

    r=(5,1,5)+T(\overrightarrow{OL}) as the question has it, almost.


    (Of course, as the negative doesnt really matter, we can just ignore it and leave the last line here with t rather than T, just as the question shows)
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Do you agree that \overrightarrow{MN}=-(\overrightarrow{OL}) ?

    So your eq. for r is just r=(5,1,5)+t(\overrightarrow{MN}) as you'd say, correct?

    But note that this is just the same as r=(5,1,5)+t(-\overrightarrow{OL})=(5,1,5)+(-t)(\overrightarrow{OL})

    Since t can be any number in \mathbb{R}, we can just get rid off the negative (as it doesn't really matter) by saying that (-t) \mapsto T, and say that:

    r=(5,1,5)+T(\overrightarrow{OL}) as the question has it, almost.


    (Of course, as the negative doesnt really matter, we can just ignore it and leave the last line here with t rather than T, just as the question shows)
    Ah, okay. I understand what you've done. So in reality, it's the t that is positive or negative (and any real number) and if it were positive, it would mean it was going in the opposite direction to if t were negative? Is that right?
    • Community Assistant
    • Welcome Squad
    Online

    20
    ReputationRep:
    (Original post by vector12)
    Ah, okay. I understand what you've done. So in reality, it's the t that is positive or negative (and any real number) and if it were positive, it would mean it was going in the opposite direction to if t were negative? Is that right?
    Yes in reality it doesn't matter because we allow t\in \mathbb{R}. Not sure what you ask in the second part there.

    It basically boils down to: if my directional vector goes the opposite way (like OL goes opposite MN), well it doesn't matter, just pick appropriate t<0 in the equation and you'll set it going in the correct direction again. The only thing that matters here is that the directional vector you pick MUST be parallel to your wanted dir. vector.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.