# Distance between planes

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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?

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?

0

reply

Report

#2

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!

0

reply

Report

#5

I agree with textbook answers. I got which is 0.291 to 3sf.

Edit. Multiply by 4 in fact and that's the correct answer.

Edit. Multiply by 4 in fact and that's the correct answer.

Last edited by B_9710; 2 years ago

0

reply

Report

#6

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

0

reply

Report

#7

(Original post by

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.

**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.

I see I used wrong equation for one of the planes.

Last edited by B_9710; 2 years ago

0

reply

Report

#8

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...

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...

1

reply

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!

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!

0

reply

Report

#10

(Original post by

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!

**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!

0

reply

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).

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).

0

reply

Report

#12

(Original post by

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).

**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).

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!

1

reply

Report

#13

(Original post by

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).

**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).

0

reply

(Original post by

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!

**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!

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top