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# C4 - 5K q8 Vectors watch

1. This seems to be slightly different than the rest of the vector questions, and it is slightly confusing.
It says:

Vectors r and s are given by

r = ai + (2a-1)j - k
s = (1-a)i + 3aj + (4a -1)k

where a is a scalar.

a) Find the values of a for which r and s are perpendicular.

I don't really know how to go about doing this...for a start there are 2 positions in the book...and I don't see how these lines can cross more than once, unless the lines don't have to cross for them to be perpendicular to eachother...

What I tried to do was to dot the directional vectors together and make that equivalent to 0 since this is when the lines would cross at 90*. But that wasn't the right asnwer...and also, how do I know that these two lines cross at one point with a 90* angle ( They don't seem to anyway)

Thank you
2. This is what i would do:

r.s = 0

( a ) . ( 1-a )
( 2a-1 ) . ( 2a )
( -1 ) . ( 4a-1)

(a)*(1-a) + (2a-1)*(2a) + (-1)*(4a-1) = 0
a - a^2 + 4a^2 - 2a - 4a + 1 = 0
3a^2 - 5a + 1 = 0

hmm does that factorise? Doubt it.

{ 5+/- SQURE ROOT [25-4*1*3] } / 6 =
= Something quiet ugly = (5/6) +/- Square root [13/36]

Now, if you plug that back into the top you'll get an answer but not very nice. Is that ANYWHERE near close? :/ I hate Vectors.
3. Wait a sec, I am trying to work this out the answers are 1 and (1/5)
EDIT argh I made a typo in the question...fixed now, but your working would now be:
r.s = 0

( a ) . ( 1-a )
( 2a-1 ) . ( 3a )
( -1 ) . ( 4a-1)

(a)*(1-a) + (2a-1)*(3a) + (-1)*(4a-1) = 0
a - a^2 + 6a^2 - 3a - 4a + 1 = 0
5a^2 - 6a + 1 = 0
Which does factorise to give the correct values it's true but put these values back into the original equation... and these points don't even happen when the lines meet...
4. well if they are perpindicular the scalar product equals 0 ..
the scalar product is i1*i2 + j1*j2 + k1*k2
soo a(1-a) + 2a(2a-1) + -1(4a-1) = 0
so we get a-a^2+4a^2-2a-4a+1 = 0
3a^2-5a+1 = 0

but there is no solution to this .. are u sure you gave me the number correctly?
5. thats what i did .. there seems to be sthg wrong ...
6. You know they cross each other at 90 degrees because the question says they are perpendicular to one another. I got the same as solly. What is the answer in the book?
7. r and s above are not equations of a line, but points (relative to the origin i presume. So you are looking for points that would be perpendicular from the origin.

So just dot prouct r and s and solve to find a. you should get a quadratic that gives two solutions.

I get, 5a^2 -6a+1=0
8. (Original post by silent ninja)
r and s above are not equations of a line, but points (relative to the origin i presume. So you are looking for points that would be perpendicular from the origin.

So just dot prouct r and s and solve to find a. you should get a quadratic that gives two solutions.

I get, 5a^2 -6a+1=0
how did u get that ??? i got 3a^2-5a+1 = 0
refer to above
9. (Original post by Sollytear)
This is what i would do:

r.s = 0

( a ) . ( 1-a )
( 2a-1 ) . ( 2a )
( -1 ) . ( 4a-1)

(a)*(1-a) + (2a-1)*(2a) + (-1)*(4a-1) = 0
a - a^2 + 4a^2 - 2a - 4a + 1 = 0
3a^2 - 5a + 1 = 0

hmm does that factorise? Doubt it.

{ 5+/- SQURE ROOT [25-4*1*3] } / 6 =
= Something quiet ugly = (5/6) +/- Square root [13/36]

Now, if you plug that back into the top you'll get an answer but not very nice. Is that ANYWHERE near close? :/ I hate Vectors.
should be 3
10. How did you get 5a^2?

a(1-a) = a - a^2
(2a-1)2a = 4a^2 - 2a
-1(4a-1) = -4a +1

add em gets you 3a^2 -5a + 1
11. o i just noticed the edit ..
12. (Original post by ma3eeni)
well if they are perpindicular the scalar product equals 0 ..
the scalar product is i1*i2 + j1*j2 + k1*k2
soo a(1-a) + 2a(2a-1) + -1(4a-1) = 0
so we get a-a^2+4a^2-2a-4a+1 = 0
3a^2-5a+1 = 0

but there is no solution to this .. are u sure you gave me the number correctly?
that should be 3a
13. Indeed...
14. he just edited the question ^
15. (Original post by silent ninja)
should be 3
HeHe, a sneeky edit i see of the question

Lets try it again then! moment while i type it all out
16. yep, i now get 1 or 1/5
17. okay you made me type this up now:

a(1-a) + (2a-1)3a - (4a-1) = 0
= a -a^2 + 6a^2 -3a -4a +1
= 5a^2 - 6a + 1
= (5a-1)(a-1)

I didnt notice the op typed it up wrong, i read the Q from the book
18. This is what i would do:

r.s = 0

( a ) . ( 1-a )
( 2a-1 ) . ( 2a )
( -1 ) . ( 4a-1)

(a)*(1-a) + (2a-1)*(3a) + (-1)*(4a-1) = 0
a - a^2 + 6a^2 - 3a - 4a + 1 = 0
5a^2 - 6a + 1 = 0

(5a-1)(a-1)
a = 1/5 OR 1

Add to start, and points are as follows:
r = ai + (2a-1)j - k

P] 1i + 1j - 1k
Q] 1/5i - 3/5j - 1k

PS: reason im doing this is so I know how to do it to
19. (Original post by Sollytear)
HeHe, a sneeky edit i see of the question

Lets try it again then! moment while i type it all out
I read it from the book, its easier compared to the formatting of maths Q's on TSR.
20. Yeah umm, sorry for the confusion...
A friends has helped me understand...basically I was treating r and s as equations of lines in vector form. Therefore, I couldn't get them to be perpendicular...I now understand that those two equations are actually points...

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