Join TSR now and get all your revision questions answeredSign up now

Vectors - Proof

    • Thread Starter
    Offline

    2
    ReputationRep:
    Hey,
    Anyone know how to crack these questions? I can't seem to get to them

    Thank you in advance
    Attached Images
     
    • Thread Starter
    Offline

    2
    ReputationRep:
    Got this for C2 - not sure if its at all right
    Attached Images
     
    Offline

    3
    (Original post by Fatts13)
    Got this for C2 - not sure if its at all right
    First line of your proof doesn't make sense with the |AC| |AC| cos theta, that would only equal |AC| |AC| if if theta were 90 degrees.

    I'd start with AC=AB+BC (note the plus sign!)

    So, AC= BC - BA (flipping the AB around)

    Now dot product.

    AC.AC = (BC-BA).(BC-BA)

    and carry on.

    For C1. Just substitute into the formula for the function.
    For part c, you want to aim to show |s(v)|^2 = |v|^2, and the result will follow.

    Use what you know about the properties of the dot product.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    First line of your proof doesn't make sense with the |AC| |AC| cos theta, that would only equal |AC| |AC| if if theta were 90 degrees.

    I'd start with AC=AB+BC (note the plus sign!)

    So, AC= BC - BA (flipping the AB around)

    Now dot product.

    AC.AC = (BC-BA).(BC-BA)

    and carry on.

    For C1. Just substitute into the formula for the function.
    For part c, you want to aim to show |s(v)|^2 = |v|^2, and the result will follow.

    Use what you know about the properties of the dot product.
    So for C2:
    AC.AC = (BC.BA).(BC-BA)
    = (BC)^2 - (BC)(BA) - (BC)(BA) + (BA)^2
    = (BC)^2 + (BA)^2 - 2(BC)(BA)
    Do I just put in cos (angle ABC) at the end? And is BA the same as AB?

    Sorry I'm asking the most stupidest questions.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Got this for C2
    Attached Images
     
    Offline

    3
    (Original post by Fatts13)
    So for C2:
    AC.AC = (BC.BA).(BC-BA)
    = (BC)^2 - (BC)(BA) - (BC)(BA) + (BA)^2
    = (BC)^2 + (BA)^2 - 2(BC)(BA)
    Do I just put in cos (angle ABC) at the end? And is BA the same as AB?

    Sorry I'm asking the most stupidest questions.
    You need to be careful which operation you're performing at any time.
    I was a bit lax in my post, however this appears to be an issue, so I'll try and be accurate.


    \vec{AC}.\vec{AC} is the dot product, and it's equal to |\vec{AC}|^2

    (BC.BA).(BC-BA) should be (\vec{BC}-\vec{BA}).(\vec{BC}-\vec{BA})

    which gives you \vec{BC}.\vec{BC} - \vec{BA}.\vec{BC} - \vex{BC}.\vec{BA} +\vec{BA}.\vec{BA}

    =\vec{BC}.\vec{BC} - \vec{BA}.\vec{BC} - \vex{BC}.\vec{BA} +\vec{AB}.\vec{AB}

    =|\vec{BC}|^2 - 2\vec{BA}.\vec{BC} +|\vec{AB}|^2

    Then from the defintion of the dot product, as you were intending:

    =|\vec{BC}|^2+|\vec{AB}|^2- 2|\vec{AB}||\vec{BC}|\cos (\angle ABC)

    Although BA is not the same as AB, as vectors they point in opposite directions and \vec{AB}=-\vec{BA}, but their lengths are the same and |\vec{BA}|=|\vec{AB}|

    You can use LaTex on TSR for mathematical symbols - see entry in Useful Resources widget to the right on the forum.

    Edit: I've assumed all the overlines are meant to be arrows, but just didn't print very well in the question.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    You need to be careful which operation you're performing at any time.
    I was a bit lax in my post, however this appears to be an issue, so I'll try and be accurate.


    \vec{AC}.\vec{AC} is the dot product, and it's equal to |\vec{AC}|^2

    (BC.BA).(BC-BA) should be (\vec{BC}-\vec{BA}).(\vec{BC}-\vec{BA})

    which gives you \vec{BC}.\vec{BC} - \vec{BA}.\vec{BC} - \vex{BC}.\vec{BA} +\vec{BA}.\vec{BA}

    =\vec{BC}.\vec{BC} - \vec{BA}.\vec{BC} - \vex{BC}.\vec{BA} +\vec{AB}.\vec{AB}

    =|\vec{BC}|^2 - 2\vec{BA}.\vec{BC} +|\vec{AB}|^2

    Then from the defintion of the dot product, as you were intending:

    =|\vec{BC}|^2+|\vec{AB}|^2- 2|\vec{AB}||\vec{BC}|\cos (\angle ABC)

    Although BA is not the same as AB, as vectors they point in opposite directions and \vec{AB}=-\vec{BA}, but their lengths are the same and |\vec{BA}|=|\vec{AB}|

    You can use LaTex on TSR for mathematical symbols - see entry in Useful Resources widget to the right on the forum.

    Edit: I've assumed all the overlines are meant to be arrows, but just didn't print very well in the question.
    I'm so sorry for having made you do all this long working out. Thank you so much, I understand it now.
    No my lecturer prefers we don't use arrows - bit weird but hey.
    Thanks for letting me know about LaTex

    Heres what i did for C1: Sorry for being a pest
    Attached Images
     
    • Thread Starter
    Offline

    2
    ReputationRep:
    Not sure about C3 :s
    Offline

    3
    (Original post by Fatts13)
    Heres what i did for C1: Sorry for being a pest
    No problem.

    Yep, C1 looks fine, though one of your "." s should be a "-" in the first line of B - very minor point, and it's obvious you meant "-".

    Geometric interpretation:
    Draw a diagram, mark a v and an e (I suggest an acute angle between them), and apply the function.

    One thing you may find helpful if you're not already aware of it, is (v.e)e is the component of v in the direction of e, when e is a unit vector.

    Answer in spoiler, BUT have a go first:
    Spoiler:
    Show



    The function s reflects v in a line perpendicular to e (I think).




    Not got time for C3 at the moment.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    No problem.

    Yep, C1 looks fine, though one of your "." s should be a "-" in the first line of B - very minor point, and it's obvious you meant "-".

    Geometric interpretation:
    Draw a diagram, mark a v and an e (I suggest an acute angle between them), and apply the function.

    One thing you may find helpful if you're not already aware of it, is (v.e)e is the component of v in the direction of e, when e is a unit vector.

    Answer in spoiler, BUT have a go first:
    Spoiler:
    Show




    The function s reflects v in a line perpendicular to e (I think).





    Not got time for C3 at the moment.
    Thank you so much. I get it now. Thank you for your time and sorry for wasting so much of your time.
    Offline

    3
    (Original post by Fatts13)
    ...
    Nope - can't see how to do C3.
    Offline

    3
    ReputationRep:
    (Original post by ghostwalker)
    Nope - can't see how to do C3.
    I think thie following works:

    Write \vec{AB} = {\bf b}, \vec{AC} = {\bf c}, \vec{AD} = \lambda {\bf b}.

    The rest is essentially mechanical - I'll put an outline behind the spoiler:

    Spoiler:
    Show


    Then |AB| |AD| |DB| becomes \lambda (1-\lambda) b^3

    and the LHS of the equality becomes

    c^2 (1-\lambda) b + \lambda b{ ({\bf b} - \bf c})^2 -  b({\bf c}-\lambda {\bf b})^2

    Expand everything out and nearly everything cancels to leave \lambda b^3 - \lambda^2 b^3[ = \lambda (1-\lambda) b^3 as required.

    (Always possible I've made a stupid mistake somewhere, but as I got the right answer I'm guessing I didn't...)



    Offline

    3
    (Original post by DFranklin)
    I think thie following works:

    Write \vec{AB} = {\bf b}, \vec{AC} = {\bf c}, \vec{AD} = \lambda {\bf b}.

    The rest is essentially mechanical - I'll put an outline behind the spoiler:

    Spoiler:
    Show





    Then |AB| |AD| |DB| becomes \lambda (1-\lambda) b^3

    and the LHS of the equality becomes

    c^2 (1-\lambda) b + \lambda b{ ({\bf b} - \bf c})^2 -  b({\bf c}-\lambda {\bf b})^2

    Expand everything out and nearly everything cancels to leave \lambda b^3 - \lambda^2 b^3[ = \lambda (1-\lambda) b^3 as required.

    (Always possible I've made a stupid mistake somewhere, but as I got the right answer I'm guessing I didn't...)






    Very nice - PRSOM.

    I don't think that method even made it to my list of things to try. Got diverted by the hint and previous working - that's my excuse, and I'm sticking to it
 
 
 
Poll
How are you feeling about your exams so far?
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

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.