# Thread: Vectors, find point to make three points a rectangle

1. ## 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
apolygise is image/link doesn't work, am posting this at school and imageshack is filtered, so I can't check...

2. 9b. $N = 3i + 4j + \left( {c - 3} \right)k$.
Sides must be parallel and equal in length.

3. Ah I see, so the vector LM has to be the same as ON?

4. 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)?

5. 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.

6. Originally Posted by LHS
I really could do with some help on 9.b
Thanks

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.

7. 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

8. 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: $(2,-3,3) \cdot (3,4,c-3) = 0$

9. 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

10. Don't worry about it, we all make mistakes! Thanks for helping though!