# Mu and lambda are parameters

1. Nov 6, 2012

### lionely

the lines l2[/SUB and l3
are given by

l1 : r = 10i + j + 9k + μ(3i+j+4k)

l2: x= (y+9)/2= (z-13)/-3

l3 r= -3i-5j-4k + λ(4i+3j+k)

where mu and lambda are parameters.

d) Show that AC=BC

A(4,-1,1) this was given, I calculated and got B as (5,-1,-2) and C (1,-2,-3)

This I what I did so far

I found AC ( -3,-1,-4)

BC(-4,-1,-1)

I also found the mods of AC and BC to see if they were equal but, they're not.

Last edited by a moderator: Feb 6, 2013
2. Nov 6, 2012

### SammyS

Staff Emeritus
Re: Vectors

It looks like you haven't given us the whole problem, although we can read between the lines to figure out most of what's missing:
Two of the lines intersect at point A, which is at (4,-1,1).

Two of the lines intersect at point B.

Two of the lines intersect at point C.​
...

Now,

Calculate vector AC and see if its magnitude matches the magnitude of either of the other two vectors.

B is at (5,1,-2) not (5,-1,-2)

Last edited: Nov 6, 2012
3. Nov 6, 2012

### lionely

Re: Vectors

it doesn't Magnitude of AC is root 26

while magnitude of B is root 30 and C is root 14.

sorry!!!
I found BC and the magnitude of it is root 26! Thank you for pointing out the mistake in the coordinates of B!!!

4. Nov 6, 2012

### SammyS

Staff Emeritus
Re: Vectors

Good deal !

5. Nov 6, 2012

### lionely

Re: Vectors

Ummm there's is one more part to the question , It says write down the coordinates of the point D on AB such that CB is perpendicular to AB.

I'm not sure what to do... do I do something with the dot product?

6. Nov 6, 2012

### SammyS

Staff Emeritus
Re: Vectors

Well, using the dot product will be a good way to check your answer, but think about the overall situation.

Triangle ABC has
length AC = length BC .​
Where should point D be on AB so that ... looks like a typo ????

7. Nov 6, 2012

### lionely

Re: Vectors

write down the coordinates of the point D on AB such that CD is perpendicular to AB.

8. Nov 7, 2012

### SammyS

Staff Emeritus
Re: Vectors

OK! That makes sense.

Now,

Triangle ABC has
length AC = length BC .​
Where should point D be on AB so that CD is perpendicular to AB.

9. Nov 7, 2012

### lionely

Re: Vectors

In the middle of AB?

10. Nov 7, 2012

### SammyS

Staff Emeritus
Re: Vectors

Yes. D is the midpoint of AB .

11. Nov 7, 2012

### lionely

Re: Vectors

LOL so wait I can find find AD then just get D and that's it?

12. Nov 7, 2012

### SammyS

Staff Emeritus
Re: Vectors

Yes. That's it.

You can check your result by seeing if the appropriate dot product is zero.

13. Nov 7, 2012

### lionely

Re: Vectors

Thank you again for the help. Homework is now finished!