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# P3 vectors!!!! watch

1. Ok, this question is from 20 May 2002 for Edexcel. It is the last Q, Q8, part d that i am stuck on. I will type the whole question out just in case some of you don't have the question.

Referred to a fixed origin, O, the points A, B and C have position vectors (9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk), where p and q are constants.

a) find in vector form an equation of the line l which passes through A and B
The answer is: 9i c- 2j +k + t(-3i +4j +5k)

Given that C lies on l
b) Find the value of p and the value of q

ans: p=6 q=11

c) calculate, in degrees, the acute angle between OC and AB

ans: 39.8 degrees to 1 dp

The point D lies on AB and is such that OD is perpidicular to AB
d) Find position vector of D

2. (Original post by Silly Sally)
Ok, this question is from 20 May 2002 for Edexcel. It is the last Q, Q8, part d that i am stuck on. I will type the whole question out just in case some of you don't have the question.

Referred to a fixed origin, O, the points A, B and C have position vectors (9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk), where p and q are constants.

a) find in vector form an equation of the line l which passes through A and B
The answer is: 9i c- 2j +k + t(-3i +4j +5k)

Given that C lies on l
b) Find the value of p and the value of q

ans: p=6 q=11

c) calculate, in degrees, the acute angle between OC and AB

ans: 39.8 degrees to 1 dp

The point D lies on AB and is such that OD is perpidicular to AB
d) Find position vector of D

As D lies on vector AB it has vector AB as its direction component. Therefore the scalar product of the direction component of AB and the whole of vector AB should equal zero, enabling you to find the variable, lamda, or t in the case you've mentioned. If you tell me what the vector equation of AB is I might be able to summarise it better.
3. (Original post by Silly Sally)
Ok, this question is from 20 May 2002 for Edexcel. It is the last Q, Q8, part d that i am stuck on. I will type the whole question out just in case some of you don't have the question.

Referred to a fixed origin, O, the points A, B and C have position vectors (9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk), where p and q are constants.

a) find in vector form an equation of the line l which passes through A and B
The answer is: 9i c- 2j +k + t(-3i +4j +5k)

Given that C lies on l
b) Find the value of p and the value of q

ans: p=6 q=11

c) calculate, in degrees, the acute angle between OC and AB

ans: 39.8 degrees to 1 dp

The point D lies on AB and is such that OD is perpidicular to AB
d) Find position vector of D

For part b...i got 19.5 degrees...not sure where you went wrong...unless its me whos wrong....

G
4. (Original post by gzftan)
For part b...i got 19.5 degrees...not sure where you went wrong...unless its me whos wrong....

G
you should get [email protected] = 70/ rt(8300) or similar
which gives @=39.8
5. (Original post by Silly Sally)
Ok, this question is from 20 May 2002 for Edexcel. It is the last Q, Q8, part d that i am stuck on. I will type the whole question out just in case some of you don't have the question.

Referred to a fixed origin, O, the points A, B and C have position vectors (9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk), where p and q are constants.

a) find in vector form an equation of the line l which passes through A and B
The answer is: 9i c- 2j +k + t(-3i +4j +5k)

Given that C lies on l
b) Find the value of p and the value of q

ans: p=6 q=11

c) calculate, in degrees, the acute angle between OC and AB

ans: 39.8 degrees to 1 dp

The point D lies on AB and is such that OD is perpidicular to AB
d) Find position vector of D

Here goes.

the equation of OD must be the same as the answer to part a. you can rearrange this as (9-3t)i + (4t-2)j + (1+5t)k. The dot product of this and thedirection of AB equals 0. giving -3(9-3t) + 4(4t-2)+ 5(1+5t) = -27 +9t +16t -8 +5 +25t = 0

50t = 30
t=3/5

substitute back into line AB for position vector

(36/5)i + (2/5)j + 4k

how does that look?

MB
6. (Original post by mockel)
you should get [email protected] = 70/ rt(8300) or similar
which gives @=39.8
This is my working...

L 1 direction vector = 4i - 5j + 3k

L 2 direction vector = i - 2j + 2k

a.b = abcos (theta)

4 + 10 + 6 = root of (50) * root of (9)* cos (theta)

cos (theta) = 20/(3*root 50)

=19.5 degrees

That's what i got...

G
7. (Original post by gzftan)
This is my working...

L 1 direction vector = 4i - 5j + 3k

L 2 direction vector = i - 2j + 2k

a.b = abcos (theta)

4 + 10 + 6 = root of (50) * root of (9)* cos (theta)

cos (theta) = 20/(3*root 50)

=19.5 degrees

That's what i got...

G
CRAP!!!!!!!!!!!!!!!!!!!!!!

Ignore me...looking at wrong question..i did get 39.8 degrees...whoops

G
8. (Original post by gzftan)
CRAP!!!!!!!!!!!!!!!!!!!!!!

Ignore me...looking at wrong question..i did get 39.8 degrees...whoops

G
yeah, i thought so....i remember doing that [email protected] = 20/3rt(50) in some other question
9. (Original post by mockel)
yeah, i thought so....i remember doing that [email protected] = 20/3rt(50) in some other question
Yep....it's cos all my work is on random pieces of paper...and my room is a mess which doesn't help....ah well....as long as i get it right in the exam i'm happy!!

G
10. btw, just incase not many people realise, there is now a 'Maths' subforum
11. (Original post by musicboy)
Here goes.

the equation of OD must be the same as the answer to part a. you can rearrange this as (9-3t)i + (4t-2)j + (1+5t)k. The dot product of this and thedirection of AB equals 0. giving -3(9-3t) + 4(4t-2)+ 5(1+5t) = -27 +9t +16t -8 +5 +25t = 0

50t = 30
t=3/5

substitute back into line AB for position vector

(36/5)i + (2/5)j + 4k

how does that look?

MB
Why in regards to the highlighted part?
12. (Original post by Silly Sally)
Why in regards to the highlighted part?
The lines are perpendicular to each other...ie. The angle between them is 90 degrees.

cos 90 = 0 ...=> a.b = 0

G
13. (Original post by Silly Sally)
Why in regards to the highlighted part?
The vectors OD and AB are perpendicular so have a dot product of one. Imaging AB being a line in 3D space and another line from the origin hiting it at 90 degrees. The shape made by the two lines is like a "T" with the bottom of the T being the origin. I hope you get it.

MB
14. (Original post by Silly Sally)
Ok, this question is from 20 May 2002 for Edexcel. It is the last Q, Q8, part d that i am stuck on. I will type the whole question out just in case some of you don't have the question.

Referred to a fixed origin, O, the points A, B and C have position vectors (9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk), where p and q are constants.

a) find in vector form an equation of the line l which passes through A and B
The answer is: 9i c- 2j +k + t(-3i +4j +5k)

Given that C lies on l
b) Find the value of p and the value of q

ans: p=6 q=11

c) calculate, in degrees, the acute angle between OC and AB

ans: 39.8 degrees to 1 dp

The point D lies on AB and is such that OD is perpidicular to AB
d) Find position vector of D

For D, give it the coordinates xi + yj + zk

OD is perpendicular to AB, so a.b = 0

You also know D lies on AB, which means the coefficients will be equal, youll get equations for x/y/z which you can sub into the equation you get from a.b = 0 Does that help?
15. Thanks everyone!!! I understand now!!!

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