Vectors - Proof Watch

Fatts13
Badges: 2
Rep:
?
#1
Report Thread starter 2 years ago
#1
Hey,
Anyone know how to crack these questions? I can't seem to get to them

Thank you in advance
Attached files
0
quote
reply
Fatts13
Badges: 2
Rep:
?
#2
Report Thread starter 2 years ago
#2
Got this for C2 - not sure if its at all right
Attached files
0
quote
reply
ghostwalker
  • Study Helper
Badges: 15
#3
Report 2 years ago
#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.
quote
reply
Fatts13
Badges: 2
Rep:
?
#4
Report Thread starter 2 years ago
#4
(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.
0
quote
reply
Fatts13
Badges: 2
Rep:
?
#5
Report Thread starter 2 years ago
#5
Got this for C2
Attached files
0
quote
reply
ghostwalker
  • Study Helper
Badges: 15
#6
Report 2 years ago
#6
(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.
quote
reply
Fatts13
Badges: 2
Rep:
?
#7
Report Thread starter 2 years ago
#7
(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 files
0
quote
reply
Fatts13
Badges: 2
Rep:
?
#8
Report Thread starter 2 years ago
#8
Not sure about C3 :s
0
quote
reply
ghostwalker
  • Study Helper
Badges: 15
#9
Report 2 years ago
#9
(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.
quote
reply
Fatts13
Badges: 2
Rep:
?
#10
Report Thread starter 2 years ago
#10
(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.
0
quote
reply
ghostwalker
  • Study Helper
Badges: 15
#11
Report 2 years ago
#11
(Original post by Fatts13)
...
Nope - can't see how to do C3.
quote
reply
DFranklin
Badges: 18
Rep:
?
#12
Report 2 years ago
#12
(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...)



0
quote
reply
ghostwalker
  • Study Helper
Badges: 15
#13
Report 2 years ago
#13
(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
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

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.

Personalise

University open days

  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
  • Bournemouth University
    Undergraduate Mini Open Day Undergraduate
    Wed, 9 Jan '19

Were you ever put in isolation at school?

Yes (221)
28.01%
No (568)
71.99%

Watched Threads

View All
Latest
My Feed