Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 114

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 127

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 136

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 137

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 141
vectors - HSN forum

Jump to content


vectors


11 replies to this topic

#1 xClairex

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 296 posts
  • Location:west lothian
  • Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
  • Gender:Female

Posted 10 March 2007 - 09:53 PM

stuck on a few questions any help will be much appreciated

1.The triangle PQR has vertices P(1,4,-1) Q(3,4,2) and R (-2,4,1). Show that PQR is:
(i)isosceles
(ii)right-angled at P

2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.





#2 dfx

    Fully Fledged Genius

  • Members
  • PipPipPipPipPipPipPip
  • 1,955 posts
  • Gender:Male

Posted 10 March 2007 - 10:48 PM

QUOTE(xClairex @ Mar 10 2007, 09:53 PM) View Post
stuck on a few questions any help will be much appreciated

1.The triangle PQR has vertices P(1,4,-1) Q(3,4,2) and R (-2,4,1). Show that PQR is:
(i)isosceles


Isosceles implies 2 lengths are the same and 2 angles are equal if that helps.

QUOTE(xClairex @ Mar 10 2007, 09:53 PM) View Post
(ii)right-angled at P


Dot product of 2 vectors extruding from the point P must be zero if they're right angled.

QUOTE(xClairex @ Mar 10 2007, 09:53 PM) View Post
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.


w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.


#3 xClairex

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 296 posts
  • Location:west lothian
  • Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
  • Gender:Female

Posted 10 March 2007 - 11:01 PM

QUOTE(dfx @ Mar 10 2007, 10:48 PM) View Post
QUOTE(xClairex @ Mar 10 2007, 09:53 PM) View Post
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.


w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.


i dont know what you are meaning for this bit :s
thnx 4 the help


#4 Steve

    Top of the Class

  • Admin
  • PipPipPipPipPip
  • 435 posts
  • Location:Edinburgh
  • Gender:Male

Posted 11 March 2007 - 12:24 PM

QUOTE(dfx @ Mar 10 2007, 10:48 PM) View Post
w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.

That's not quite right.

For any vector u, \mathbf{u}.\mathbf{u}=|\mathbf{u}|^2.

There is a similar example on page 124 of the Vectors notes. Does this help?
HSN contribute: Help the site grow!

Looking for a Maths tutor in West Lothian? Just PM me!

#5 xClairex

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 296 posts
  • Location:west lothian
  • Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
  • Gender:Female

Posted 11 March 2007 - 05:29 PM

i still dont really understand.heres what i think im supposed to do but im not sure

if u.u=u

then w.w=w = (2u+1/2v) = (2u+1/2v)(2u+1/2v) = 4u + 2u1/2v + 2u1/2v + 1/4v
= 4u + 4uv + 1/4v

i dont know where to go from here

is this right or completly wrong?


#6 The Wedge Effect

    HSN Legend

  • Members
  • PipPipPipPipPipPipPipPip
  • 2,477 posts
  • Location:Paisley
  • Interests:I'm me, I guess. I wear glasses, have short spiked hair, about 5ft 9 (I think). I left fifth year of school in 2005 and is currently unemployed. I used to go to Strathclyde University to do MEng Chemical Engineering, but I hated the course, so I quit. I've started doing MSc Mathematics, from September 2006, as I have a keen (almost unhealthy) interest in Mathematics, and its applications.<br /><br />The grades I achieved in the exams:<br /><br />1 - Standard Grade Mathematics<br />1 - Standard Grade Physics<br />1 - Standard Grade Chemistry<br />1 - Standard Grade Computing Studies<br />1 - Standard Grade Craft and Design<br />2 - Standard Grade English<br />2 - Standard Grade Geography <br /><br />A (Band 1) - Higher English<br />A (Band 1) - Higher Mathematics<br />A (Band 2) - Higher Chemistry<br />B (Band 3) - Higher Physics<br /><br />Anything else ya wanna know about me, PM me or add me to your MSN contact list and chat to me there. :P
  • Gender:Male

Posted 11 March 2007 - 09:37 PM

Completely wrong. These lines on either side means the magnitude, or length of w, which is \left|{\bf w}\right|=\sqrt{x^2+y^2+z^2}, where x, y and z are the coefficients (number the variable is multiplied by) of the vector.

#7 dfx

    Fully Fledged Genius

  • Members
  • PipPipPipPipPipPipPip
  • 1,955 posts
  • Gender:Male

Posted 11 March 2007 - 10:12 PM

QUOTE(Steve @ Mar 11 2007, 12:24 PM) View Post
QUOTE(dfx @ Mar 10 2007, 10:48 PM) View Post
w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.

That's not quite right.



Hmm got abit carried away with the us and vs as is and js. Sorry about that.

#8 xClairex

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 296 posts
  • Location:west lothian
  • Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
  • Gender:Female

Posted 13 March 2007 - 04:33 PM

i am grateful for the help i just dont understand

i have w.w but i dont know how to find the magnitude of w as we dont have the the components of U or V ?

does anyone know how i find them?


#9 George

    Child Prodigy

  • Admin
  • PipPipPipPipPipPip
  • 720 posts
  • Location:West Lothian
  • Gender:Male

Posted 13 March 2007 - 05:24 PM

You'll need to use the fact that Steve mentioned earlier:

QUOTE(Steve @ Mar 11 2007, 12:24 PM) View Post
For any vector u, \mathbf{u}.\mathbf{u}=|\mathbf{u}|^2.


So in this case, \mathbf{w}.\mathbf{w}=|\mathbf{w}|^2, meaning you can find the magnitude of w by taking the square root of w.w.

#10 xClairex

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 296 posts
  • Location:west lothian
  • Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
  • Gender:Female

Posted 13 March 2007 - 05:55 PM

i dont think i have got w.w right i thought i would need to find u and then v so tht i get an actual number instead of equation for the magnitude of w.
me and my friend were working together and we got w.w to be 4u + 4uv + 1/4v i think someone said before that this was wrong


#11 George

    Child Prodigy

  • Admin
  • PipPipPipPipPipPip
  • 720 posts
  • Location:West Lothian
  • Gender:Male

Posted 14 March 2007 - 03:26 PM

QUOTE(xClairex @ Mar 13 2007, 05:55 PM) View Post
i dont think i have got w.w right i thought i would need to find u and then v so tht i get an actual number instead of equation for the magnitude of w.
me and my friend were working together and we got w.w to be 4u + 4uv + 1/4v i think someone said before that this was wrong

Yes, your answer isn't quite right, but you have the right idea.


QUOTE(xClairex @ Mar 10 2007, 09:53 PM) View Post
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.


The way to do this is to use the properties of the scalar product, so:



\begin{align*}
\mathbf{w}.\mathbf{w} &= (2\mathbf{u}+\frac12\mathbf{v}).(2\mathbf{u}+\frac12\mathbf{v}) \\
&= 2\mathbf{u}.2\mathbf{u} + 2\mathbf{u}.\frac12\mathbf{v} + 2\mathbf{u}.\frac12\mathbf{v} + \frac12\mathbf{v}.\frac12\mathbf{v} \\
&= 4\mathbf{u}.\mathbf{u} + 2\mathbf{u}.\mathbf{v} + \frac14\mathbf{v}.\mathbf{v}
\end{align*}

So now we can get a value for \mathbf{w}.\mathbf{w} if we can find values for each of those three terms in the last line. Can you work these out from the information in the question?

#12 xClairex

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 296 posts
  • Location:west lothian
  • Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
  • Gender:Female

Posted 14 March 2007 - 11:08 PM

aahh i get it now thank you






1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users