TAEuler
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Find the perpendicular distance between the pair of planes:

Plane 1: 4x + y - 2z = 2

Plane 2: -6x - 1.5y + 3z = 5

Give your answer to 3 significant figures:

the direction vector of plane 2 = -1.5 x the direction vector of plane 1. Therefore the planes are parallel.

Plane 1 contains the point A (0, 2, 0)

the line that passes through A and plane 2 is r = 2j + s(-6i - 1.5j + 3k)
= (-6si + (2 - 1.5s)j + 3sk)

at the point of intersection where the line and plane intersect, point Q:

(-6si + (2 - 1.5s)j + 3sk) . (-6i - 1.5j + 3k) = 5

36s - 3 + 9s/4 + 9s = 5

189s/4 = 8

s = 32/189

Vector OQ = -(64/63)i + (110/63)j + (32/63)k

Therefore vector AQ = -(64/63)i -(16/63)j + (32/63)k

The perpendicular distance = lAQl = 1.16

Textbook says 0.291, mistake in textbook or have I gone wrong?
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√-1
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Im not an expert at vectors, actually my least favourite chapter in further maths but ill give it a go and see what i get!
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username3249896
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I get 1.16 too

+ wolfram agrees
https://www.wolframalpha.com/input/?...y+%2B+3z+%3D+5
Last edited by username3249896; 2 years ago
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√-1
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thank god I can't figure it out I'm too tired
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B_9710
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I agree with textbook answers. I got  \frac{4}{63} \sqrt{21} which is 0.291 to 3sf.
Edit. Multiply by 4 in fact and that's the correct answer.
Last edited by B_9710; 2 years ago
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ghostwalker
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(Original post by B_9710)
I agree with textbook answers. I got  \frac{4}{63} \sqrt{21} which is 0.291 to 3sf.
Textbook is incorrect.

If you used the standard formula for the distance of a plane from the origin and subtracted the two values, you will get the book's answer, however that formula gives an unsigned quantity - modulus signs are required.

It can be adapted to provide a neat solution in this case though.
Last edited by ghostwalker; 2 years ago
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B_9710
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(Original post by ghostwalker)
Textbook is incorrect.

If you used the standard formula for the distance of a plane from the origin and subtracted the two values, you will get the book's answer, however that formula gives an unsigned quantity - modulus signs are required.

It can be adapted to provide a neat solution in this case though.
Are you sure? I didn't use a standard formula or anything just used a bit of intuition and can't see how it couldn't be incorrect.

I see I used wrong equation for one of the planes.
Last edited by B_9710; 2 years ago
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begbie68
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transforming the 2nd eqn to the same format as the 1st
gives
4x + y - 2z = -10/3

|-10/3 - 2| = 16/3
but there's a factor of size sqrt(4^2 + 1^2 + 2^2) = root21

hence, dist btwn planes = 16/(3root21) = 1.163828748...
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TAEuler
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Wow! This really is laughable.

This question is part of the Oxford Year 2 Textbook for Further Maths. It has 3 pairs of planes for question 2, the question has the title "Find the perpendicular distance between each of these pairs of planes. Give your answers to 3 significant figures."

I believe all 3 pairs have an incorrect solution in the back of the textbook, if someone would be willing to confirm...

First Part:
Plane 1: 3x - y + 9z = 15
Plane 2: 6x - 2y + 18z = 3

I believe the answer is 1.42 to 3 significant figures, the textbook says 1.49 (I checked on Maths Stack Exchange and they seem to believe the textbook is incorrect).

Second part:
What I've listed here

Third part:
Okay so the equation given for the first plane is x + 3y - 5z = 12 and for the second plane 2x - 6y + 10z = -13

These planes aren't perpendicular, so I'm going to assume the first plane was instead meant to be -x + 3y - 5z = 12.

Even in this case the textbook's answer for the distance is different, 3.13 units is its answer, whereas wolfram says 0.93!!

This is actually laughably bad from the textbook. 3 incorrect answers in a row I believe!
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RDKGames
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(Original post by TAEuler)
Third part:
Okay so the equation given for the first plane is x + 3y - 5z = 12 and for the second plane 2x - 6y + 10z = -13

These planes aren't perpendicular, so I'm going to assume the first plane was instead meant to be -x + 3y - 5z = 12.

Even in this case the textbook's answer for the distance is different, 3.13 units is its answer, whereas wolfram says 0.93!!

This is actually laughably bad from the textbook. 3 incorrect answers in a row I believe!
If you change the RHS to -12 then you get the book's answer.
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TAEuler
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Hopelessly lost on this question:

The lines r = (i - 2k) + s(5i - 2j - 4k) and r = (16i + 10j + k) + t(-i + 6j + 6k) intersect the plane r . (i -3j + 2k) = 9 at the points A and B respectively. Find the area of triangle OAB:

Intersection of line 1 and the plane:

r = (1 +5s)i - 2sj +(-2-4s)k

((1 +5s)i - 2sj +(-2-4s)k) . (i - 3j + 2k) = 9

1 + 5s + 6s - 4 - 8s = 9

s = 4

A is (21, -8, -18)

Intersection of line 2 and plane:

r = (16 - t)i + (10 + 6t)j + (1+ 6t)k

((16 - t)i + (10 + 6t)j + (1+ 6t)k) . (i - 3j + 2k) = 9

16 - t - 30 - 18t + 2 + 12t = 9
-7t - 12 = 9
-7t = 21
t = -3

Therefore B is (19, -8, -17)

For the area I think you should be able to do the cross product on both vectors and then find the magnitude of that and then divide by 2 to find the area, but this gives a drastically different answer to the one found in the textbook (269 I believe).
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RDKGames
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(Original post by TAEuler)
For the area I think you should be able to do the cross product on both vectors and then find the magnitude of that and then divide by 2 to find the area, but this gives a drastically different answer to the one found in the textbook (269 I believe).
Valid method, the area is not 269.

I think at this stage you may as well don't check with their answers. It gets tiring for us to point out the book is obviously wrong over and over again.

Just find a different resource with correct answers!
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SamBandara
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(Original post by TAEuler)
Hopelessly lost on this question:

The lines r = (i - 2k) + s(5i - 2j - 4k) and r = (16i + 10j + k) + t(-i + 6j + 6k) intersect the plane r . (i -3j + 2k) = 9 at the points A and B respectively. Find the area of triangle OAB:

Intersection of line 1 and the plane:

r = (1 +5s)i - 2sj +(-2-4s)k

((1 +5s)i - 2sj +(-2-4s)k) . (i - 3j + 2k) = 9

1 + 5s + 6s - 4 - 8s = 9

s = 4

A is (21, -8, -18)

Intersection of line 2 and plane:

r = (16 - t)i + (10 + 6t)j + (1+ 6t)k

((16 - t)i + (10 + 6t)j + (1+ 6t)k) . (i - 3j + 2k) = 9

16 - t - 30 - 18t + 2 + 12t = 9
-7t - 12 = 9
-7t = 21
t = -3

Therefore B is (19, -8, -17)

For the area I think you should be able to do the cross product on both vectors and then find the magnitude of that and then divide by 2 to find the area, but this gives a drastically different answer to the one found in the textbook (269 I believe).
Maybe it's just a printing mistake in the question, like RDKGames said. It happens often.
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TAEuler
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(Original post by RDKGames)
Valid method, the area is not 269.

I think at this stage you may as well don't check with their answers. It gets tiring for us to point out the book is obviously wrong over and over again.

Just find a different resource with correct answers!
Answer I got is 11.7?
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RDKGames
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(Original post by TAEuler)
Answer I got is 11.7?
Sounds good.
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