P3 vector quesitons Watch

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Freddy C
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#1
Report Thread starter 13 years ago
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Got stuck on these:

Q) The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k. The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k. The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point of PQ.

Q) referred to an origin O, the points A, B, C and D are (3,1,-1) , (6,7,8) , (2,5,0) and (0,7,-2) respectively.

calculate the size of angle ACD to the nearest degree.

For this question above, I keep getting the acute angle of the 2 vectors instead of it's obtuse one, grrrr...

Thanks for any answers
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Fermat
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Q1)
r1 = (1,1,2) + λ(2,1,-1)
r2 = (1,1,4) + μ(1,1,-2)


let PQ be the vector (x,y,z)

Then PQ.r1 = 0
(x,y,z).(2,1,-1) = 0
2x + y - z = 0
===========

PQ.r2 = 0
(x,y,z).(1,1,-2) = 0
x + y - 2z = 0
==========

let z = t, then
2x + y = t
x + y = -2t
giving,
x = -t
y = 3t
So the vector PQ is (-t, 3t, t)
PQ = t(-1,3,1)
===========

let r3 be the line through P and Q
r3 = Q + t(-1,3,1)
=============

Q is a point on the line r2 for some value of μ, and r2 can be rewritten as
r2 = (1+μ)i + (1+μ)j + (4 - 2μ)k
r2 = s.i + s.j + (6 - 2s)k = Q ----------------(1)
====================

P is a point on the line r1 for some value of λ, and r1 can be rewritten as
r1 = (1 + 2λ)i + (1 + λ)j + (2 - λ)k = P --------(2)
==========================

P is also a point on the line r3, so can be written as
P = Q + t(-1,3,1)

substituting for (1) and (2),
(1 + 2λ)i + (1 + λ)j + (2 - λ)k = s.i + s.j + (6 - 2s)k + t(-1,3,1)
(1 + 2λ)i + (1 + λ)j + (2 - λ)k = (s - t)i + (s + 3t)j + (6 - 2s + t)k

equating coeffts of i,j,k,
1 + 2λ = s -t
1 + λ = s + 3t
2 - λ = 6 - 2s + t

we end up with,
t = -2/11
λ = 8/11
s = 25/11
=======

Q = s.i + s.j + (6 - 2s)k
Q = (25/11)i + (25/11)j + (16/11)k
=========================

P = (1 + 2λ)i + (1 + λ)j + (2 - λ)k
P = (27/11)i + (19/11)j + (14/11)k
========================

p.s. check my arithmetic - I've made a couple of mistakes so far.

r3 = Q + t(-1,3,1)
r3 = (25/11)i + (25/11)j + (16/11)k + t(-1,3,1)
================================ ===

Edit: correected a couple of typos. Finally ended up with the book answer.
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Fermat
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(Original post by Freddy C)
...

Q) referred to an origin O, the points A, B, C and D are (3,1,-1) , (6,7,8) , (2,5,0) and (0,7,-2) respectively.

calculate the size of angle ACD to the nearest degree.

For this question above, I keep getting the acute angle of the 2 vectors instead of it's obtuse one, grrrr...

Thanks for any answers
I thnk you may have been taking the dot product of AC and CD, rather than CA and CD.
The two vectors should have their bases together and their direction arrows pointing away from each other, rather than following each other around, as in a vector diagram.

A(3,1,-1), C(2,5,0), D(0,7,-2)

CA = A - C
CA = (1, -4, -1)

|CA| = √(1 + 16 + 1)
|CA| = √18

CD = D - C
CD = (-2, 2, -2)
|CD| = √(4 + 4 + 4)
|CD| = √12

CA.CD = (1, -4, -1).(-2, 2, -2)
CA.CD = -2 -8 + 2
CA.CD = -8

CA.CD = |CA||CS|[email protected]
-8 = √18.√[email protected]
[email protected] = -8/√18.√12
@ = 180 - 57
@ = 123
======
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Freddy C
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(Original post by Fermat)
P is also a point on the line r3, so can be written as
P = Q + t(-3,1,1)

sunstituting for (1) and (2),
(1 + 2λ)i + (1 + λ)j + (2 - λ)k = s.i + s.j + (6 - 2s)k + t(-3,1,1)
(1 + 2λ)i + (1 + λ)j + (2 - λ)k = (s - t)i + (s + 3t)j + (6 - 2s + t)k

equating coeffts of i,j,k,
1 + 2λ = s -t
1 + λ = s + 3t
2 - λ = 6 - 2s + t

we end up with,
t = -2/11
λ = 8/11
s = 25/11
=======

Q = s.i + s.j + (2 - 2s)k
Q = (25/11)i + (25/11)j - (28/11)k
=======================

P = (1 + 2λ)i + (1 + λ)j + (2 - λ)k
P = (27/11)i + (19/11)j + (14/11)k
========================

p.s. check my arithmetic - I've made a couple of mistakes so far.
I'm fine until the start of the quote above, where did P = Q + t(-3,1,1) come from?

I would have thought it was P = Q + t(-1,3,1).

The answer in the book is r=(25/11)i+(25/11)j+(16/11)k + λ(-i+3j+k)

On the second one, yea you are right, thank you :)
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Fermat
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(Original post by Freddy C)
I'm fine until the start of the quote above, where did P = Q + t(-3,1,1) come from?

I would have thought it was P = Q + t(-1,3,1).

The answer in the book is r=(25/11)i+(25/11)j+(16/11)k + λ(-i+3j+k)

On the second one, yea you are right, thank you
Yeh, sorry. That was a typo. The line after it was ok.
Corrected that typo and another one in my original post.
Also showed the final answer as from the book.
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Freddy C
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Thanks for the correction, one last question

with r3 = Q + t(-1,3,1), can you swap Q for P?

if not, then any reasons why Q was choosen?
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Fermat
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(Original post by Freddy C)
Thanks for the correction, one last question

with r3 = Q + t(-1,3,1), can you swap Q for P?

if not, then any reasons why Q was choosen?
you can swap P for Q.

a line is given by,

l = position vector (point) plus parameter(e.g. lambda) times vector

so you can change your position vector from Q to P. It just means a change in the value of t (or,lambda).
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