# A Question For Rahayden

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Hi! I Was Just Wondering If You Had Looked At Jan 2004 Edexcel Paper P3. I Wanted To Ask Help On The Last Question Of The Paper. Thanks

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#2

Do you mean the vectors question if so here it is:

L1 : r = i + 3j + 5k + lamda(i + 2j - k)

L2 : r = -2i + 3j - 4k + mew(2i + j + 4k)

a) Show L1 and L2 intersect and then give the coordinates of this intersection Q.

b) Show l1 and l2 are perpendicular

c) Point P is on L1 with x-coordinate 3. Point R is on L2 with x-coordinate 4 .

Find in the simplest form the exact area of Triangle PQR.

I have offered my solution to this on the other thread but would be interested to see other solutions and thoughts. Mensandan

L1 : r = i + 3j + 5k + lamda(i + 2j - k)

L2 : r = -2i + 3j - 4k + mew(2i + j + 4k)

a) Show L1 and L2 intersect and then give the coordinates of this intersection Q.

b) Show l1 and l2 are perpendicular

c) Point P is on L1 with x-coordinate 3. Point R is on L2 with x-coordinate 4 .

Find in the simplest form the exact area of Triangle PQR.

I have offered my solution to this on the other thread but would be interested to see other solutions and thoughts. Mensandan

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#3

(Original post by

Hi! I Was Just Wondering If You Had Looked At Jan 2004 Edexcel Paper P3. I Wanted To Ask Help On The Last Question Of The Paper. Thanks

**thecaped**)Hi! I Was Just Wondering If You Had Looked At Jan 2004 Edexcel Paper P3. I Wanted To Ask Help On The Last Question Of The Paper. Thanks

Or at least put this on the academic subforum.

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#4

Good point, keep forgetting to do that. Will try harder in future. Good idea about the PM though, cheers dude.

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#5

Forgive me if I make any mistakes, as I'm tired and doing this as I type. Also, I use the convention a = lambda, b = mew.

To find where they intersect, solve the system of linear equations (or just spot appropriate values of a and b by sight):

a - 2b = -3

2a - b = 0

a + 4b = 9

This yields one solution a = 1, b = 2, and the point of intersection Q = (2, 5, 4).

The dot product of their direction vectors is:

(1, 2, -1).(2, 1, 4) = 2 + 2 - 4= 0, which implies that the angle between the lines has cosine of 0, so they are perpendicular.

We want the area of the triangle defined by the three points P, Q and R, where:

P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

As long as I haven't made any calculation errors.

I hope this helps,

(Original post by

Do you mean the vectors question if so here it is:

L1 : r = i + 3j + 5k + lamda(i + 2j - k)

L2 : r = -2i + 3j - 4k + mew(2i + j + 4k)

a) Show L1 and L2 intersect and then give the coordinates of this intersection Q.

**mensandan**)Do you mean the vectors question if so here it is:

L1 : r = i + 3j + 5k + lamda(i + 2j - k)

L2 : r = -2i + 3j - 4k + mew(2i + j + 4k)

a) Show L1 and L2 intersect and then give the coordinates of this intersection Q.

a - 2b = -3

2a - b = 0

a + 4b = 9

This yields one solution a = 1, b = 2, and the point of intersection Q = (2, 5, 4).

(Original post by

b) Show l1 and l2 are perpendicular

**mensandan**)b) Show l1 and l2 are perpendicular

(1, 2, -1).(2, 1, 4) = 2 + 2 - 4= 0, which implies that the angle between the lines has cosine of 0, so they are perpendicular.

(Original post by

c) Point P is on L1 with x-coordinate 3. Point R is on L2 with x-coordinate 4 .

Find in the simplest form the exact area of Triangle PQR.

**mensandan**)c) Point P is on L1 with x-coordinate 3. Point R is on L2 with x-coordinate 4 .

Find in the simplest form the exact area of Triangle PQR.

P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

As long as I haven't made any calculation errors.

I hope this helps,

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#6

(Original post by

Forgive me if I make any mistakes, as I'm tired and doing this as I type. Also, I use the convention a = lambda, b = mew.

To find where they intersect, solve the system of linear equations (or just spot appropriate values of a and b by sight):

a - 2b = -3

2a - b = 0

a + 4b = 9

This yields one solution a = 1, b = 2, and the point of intersection Q = (2, 5, 4).

The dot product of their direction vectors is:

(1, 2, -1).(2, 1, 4) = 2 + 2 - 4= 0, which implies that the angle between the lines has cosine of 0, so they are perpendicular.

We want the area of the triangle defined by the three points P, Q and R, where:

P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

As long as I haven't made any calculation errors.

I hope this helps,

**rahaydenuk**)Forgive me if I make any mistakes, as I'm tired and doing this as I type. Also, I use the convention a = lambda, b = mew.

To find where they intersect, solve the system of linear equations (or just spot appropriate values of a and b by sight):

a - 2b = -3

2a - b = 0

a + 4b = 9

This yields one solution a = 1, b = 2, and the point of intersection Q = (2, 5, 4).

The dot product of their direction vectors is:

(1, 2, -1).(2, 1, 4) = 2 + 2 - 4= 0, which implies that the angle between the lines has cosine of 0, so they are perpendicular.

We want the area of the triangle defined by the three points P, Q and R, where:

P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

As long as I haven't made any calculation errors.

I hope this helps,

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#7

(Original post by

We want the area of the triangle defined by the three points P, Q and R, where:

P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

**rahaydenuk**)We want the area of the triangle defined by the three points P, Q and R, where:

P = (3, 7, 3), Q = (2, 5, 4) and R = (4, 6, 8)

If we now perform a rigid transformation of the plane (so it does not affect the triangle's shape and thus its area), by translating the plane such that Q is at the origin, we have reduced the problem to finding the area of the triangle defined by the points:

O (the origin), P' = (1, 2, -1) and R' = (2, 1, 4)

It's easy to prove that the area of a triangle with one corner at the origin and the other two corners given by position vectors t and s is (1/2)|t x s|, so the area of our triangle is:

Area = (1/2) | (1, 2, -1) x (2, 1, 4) | = (1/2)Sqrt(126) = (3/2)Sqrt(14)

*as long as*you know about cross/vector products and determinants; but I don't think those come along until P6!

Therefore for the benefit of those who did not have knowledge of P6 methods in their P3 exam I will provide an alternative solution for this part of the question.

We know that the lines L1 and L2 are perpendicular, and that the points P (3,7,3) and R (4,6,8) lie on lines L1 and L2 respectively. We also know that Q (2,5,4) is the point of intersection between these two lines. Hence we are dealing with a right angled triangle and can calculate the area by using the GCSE formula 1/2 * base * height. We can calculate the length of the base and height by pythagoras. Thus...

height = sqrt((3-2)^2+(7-5)^2+(3-4)^2) = sqrt6

base = sqrt((2-4)^2+(5-6)^2+(4-8)^2) = sqrt21

=> Area = 1/2 * sqrt6 * sqrt21 = 1/2 * sqrt126 = 3/2 * sqrt14

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The question about solving the partial fraction differential equation, what was the final answer, please?

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#9

(Original post by

The question about solving the partial fraction differential equation, what was the final answer, please?

**thecaped**)The question about solving the partial fraction differential equation, what was the final answer, please?

y = 4/(2x-3) - 3/(x+1)

Integrating you get:

Lny = 2Ln(2x-3) - 3Ln(x+1) + c

Now it said y=4 when x=2 so:

ln4 = ln 1 - ln 27 + c

So C = ln 4 + ln 27 = ln 108

Final answer

ln y = 2ln(2x-3) - 3ln(x+1) + ln108

ln y = 108 ((2x-3)^2/(x+1)^3)

Or

y=108((2x-3)^2/(x+1)^3)

This seems to be correct though I think I messed up the constant part in the exam. Never mind.

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#10

(Original post by

The Partial Fraction was:

y = 4/(2x-3) - 3/(x+1)

Integrating you get:

Lny = 2Ln(2x-3) - 3Ln(x+1) + c

Now it said y=4 when x=2 so:

ln4 = ln 1 - ln 27 + c

So C = ln 4 + ln 27 = ln 108

Final answer

ln y = 2ln(2x-3) - 3ln(x+1) + ln108

ln y = 108 ((2x-3)^2/(x+1)^3)

Or

y=108((2x-3)^2/(x+1)^3)

This seems to be correct though I think I messed up theconstant part in the exam. Never mind.

**mensandan**)The Partial Fraction was:

y = 4/(2x-3) - 3/(x+1)

Integrating you get:

Lny = 2Ln(2x-3) - 3Ln(x+1) + c

Now it said y=4 when x=2 so:

ln4 = ln 1 - ln 27 + c

So C = ln 4 + ln 27 = ln 108

Final answer

ln y = 2ln(2x-3) - 3ln(x+1) + ln108

ln y = 108 ((2x-3)^2/(x+1)^3)

Or

y=108((2x-3)^2/(x+1)^3)

This seems to be correct though I think I messed up theconstant part in the exam. Never mind.

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#11

(Original post by

Of course your method is completely sound,

**Darkness**)Of course your method is completely sound,

*as long as*you know about cross/vector products and determinants; but I don't think those come along until P6!
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#12

(Original post by

I didn't use determinants in that solution.

**rahaydenuk**)I didn't use determinants in that solution.

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#13

(Original post by

simplay the best mathematician i know.

**bono**)simplay the best mathematician i know.

but the method as pretty long and intimidating !!!

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#14

(Original post by

wow. another good mathematician on the forum....

**bono**)wow. another good mathematician on the forum....

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