# Vectors, find point to make three points a rectangle

• March 16th 2009, 06:57 AM
LHS
Vectors, find point to make three points a rectangle
I really could do with some help on 9.b
Thanks :)
http://img7.imageshack.us/img7/5531/scan0004r.jpg
http://img7.imageshack.us/img7/5531/scan0004r.jpg
apolygise is image/link doesn't work, am posting this at school and imageshack is filtered, so I can't check...
• March 16th 2009, 07:36 AM
Plato
9b. $N = 3i + 4j + \left( {c - 3} \right)k$.
Sides must be parallel and equal in length.
• March 16th 2009, 11:28 AM
LHS
Ah I see, so the vector LM has to be the same as ON?
• March 16th 2009, 11:34 AM
Plato
Quote:

Originally Posted by LHS
Ah I see, so the vector LM has to be the same as ON?

Did you find c is part (a)?
• March 16th 2009, 11:58 AM
LHS
Yes, C=5, and now we have point N part c should be very simple, thank you for your help!
I kept trying to have ON and MN dotted = to zero to create simultaneous equations, with equations for the line MN's length equalling the length of OL.
It failed quite superbly, I greatly over complicated matters, would have been useful to know how many marks this question was.
• March 16th 2009, 12:18 PM
earboth
Quote:

Originally Posted by LHS
I really could do with some help on 9.b
Thanks :)
http://img7.imageshack.us/img7/5531/scan0004r.jpg
http://img7.imageshack.us/img7/5531/scan0004r.jpg
apolygise is image/link doesn't work, am posting this at school and imageshack is filtered, so I can't check...

If I didn't misread the question completely then the vectors $\overrightarrow{OL}$ and $\overrightarrow{OM}$ must be perpendicular. Therefore:

$(2,-3,3) \cdot (5,1,c) = 0~\implies~c=-\dfrac73$

The position vector of N is:

$\overrightarrow{ON} = \overrightarrow{OL} + \overrightarrow{OM} = \left(7,-2,\frac23 \right)$

The equation of the line is then:

$\vec r = \overrightarrow{OM} + t \cdot (\overrightarrow{ON} - \overrightarrow{OM})$

Plug in the vectors you know to get the equation of the line.
• March 16th 2009, 12:26 PM
LHS
I'm pretty sure that the lines LM and OL are perpendicular.
(3i 4j (c-3)k).(2i -3j 3k)=0
6-12+3c-9=0
3c=15
c=5
• March 16th 2009, 12:34 PM
Plato
Quote:

Originally Posted by earboth
If I didn't misread the question completely then the vectors $\color{red}\overrightarrow{OL}$ and $\color{red}\overrightarrow{OM}$ must be perpendicular.

Due to the order of the vertices $OLMN$ we would have $\overrightarrow{OL}$ and $\overrightarrow{LM}$ must be perpendicular.
Therefore: $(2,-3,3) \cdot (3,4,c-3) = 0$
• March 16th 2009, 12:35 PM
earboth
Quote:

Originally Posted by LHS
I'm pretty sure that the lines LM and OL are perpendicular.
(3i 4j (c-3)k).(2i -3j 3k)=0
6-12+3c-9=0
3c=15
c=5

If the points are in clockwise order then you are right. Sorry for the confusion (Thinking)
• March 16th 2009, 12:39 PM
LHS
Don't worry about it, we all make mistakes! Thanks for helping though!