Fp3 vector dot product help!

So hey this is second question of mine I'd ask today if that's fine. So I got a problem when I encountered a Q bout shortest distance between planes, which I applied the equation r.n=d where d is the distance from plane to origin.

Above is the edexcel fp3 solution to the question(b part) where to find shortest distance between 2 parallel planes.

In the second plane where r.(6i+8j-24k)=52 becomes divided by (-2) gives of a negative distance. What I did was directly make (6i+8j-24k) a unit vector by the usual method without involving to -ve signs, that will give off a +ve distance(where d=2 and NOT -2) and hence deduction makes the answer as 3units. Can somebody explain this to me..

F3C5FS10.jpg111.3KB

Reply 1

ghostwalker

Original post by Hsakbo

What I did was directly make (6i+8j-24k) a unit vector by the usual method without involving to -ve signs, that will give off a +ve distance(where d=2 and NOT -2) and hence deduction makes the answer as 3units. Can somebody explain this to me..

Can you show your working for this.

Reply 2

Hsakbo

Original post by ghostwalker

Can you show your working for this.

It is the same as the one you see in pic I clipped(B Part), and like i mentioned earlier it is just about that last part on the 2nd plane where they divide r.(6i+8j-24k)=52 by a -2.

btw sorry for the big picture..

(edited 9 years ago)

Reply 3

davros

Original post by Hsakbo

n the second plane where r.(6i+8j-24k)=52 becomes divided by (-2) gives of a negative distance. What I did was directly make (6i+8j-24k) a unit vector by the usual method without involving to -ve signs, that will give off a +ve distance(where d=2 and NOT -2) and hence deduction makes the answer as 3units. Can somebody explain this to me..

What does this mean? Normalising the vector won't affect the sign of the answer, so you've gone wrong somewhere in your working :smile:

Reply 4

Hsakbo

Ok here's my working from the part where I deviated from the answer:

r.(6i+8j-24k) = 52
divide by 26 to make (6i+8j-24k) an unit vector
r.(1/26)(6i+8j-24k) = 52/26 = 2

now you see the prob? the book answer for this part is -2

btw this forum is filled with bugggs...

Reply 5

ghostwalker

Original post by Hsakbo

OK, makes sense now.

The distance to the origin depends on which direction you take the normal vector.

If you have a normal vector of n, say, giving a distance of 6, say, then choosing the normal vector of -n would give a distance of -6.

In your second working you chose a normal in the opposite direction to that of your first working, hence the problem.

Reply 6

Hsakbo

Original post by ghostwalker

So that's the problem? You have to have both normal vectors at same K vectors or something??

Reply 7

ghostwalker

Original post by Hsakbo

So that's the problem? You have to have both normal vectors at same K vectors or something??

If there's a k component, they yes, otherwise match one of the other components.

In the edexcel solution the normal vector used is reduced to (-3i-4j+12k) for both planes.

Reply 8

Hsakbo

Original post by ghostwalker

If there's a k component, they yes, otherwise match one of the other components.

In the edexcel solution the normal vector used is reduced to (-3i-4j+12k) for both planes.

I see, thanks ghostwalker! This wasn't mentioned in the book. :smile:

Reply 9

Gome44

Original post by Hsakbo

I see, thanks ghostwalker! This wasn't mentioned in the book. :smile:

I am sitting fp3 this year and found this formula in the formula book which they don't tell you in the textbook. Just thought I'd share it with you ImageUploadedByStudent Room1425133646.508735.jpg

ImageUploadedByStudent Room1425133646.508735.jpg

Reply 10

davros

Original post by Gome44

I am sitting fp3 this year and found this formula in the formula book which they don't tell you in the textbook. Just thought I'd share it with you ImageUploadedByStudent Room1425133646.508735.jpg

If you're referring to the formula right at the bottom, note that this appears to give you the magnitude of the distance i.e. it won't necessarily help in the situation where you have points on opposite sides of a plane :smile:

Reply 11

Gome44

Original post by davros

Ah yes, I didn't spot that. But surely if you just ignore the magnitude sign, the formula gives you 'displacement' rather than distance, which tells you which side of the plane the point is on?

Reply 12

davros

Original post by Gome44

Ah yes, I didn't spot that. But surely if you just ignore the magnitude sign, the formula gives you 'displacement' rather than distance, which tells you which side of the plane the point is on?

I think that's correct because the thing inside the modulus looks just like the result of a dot product. The issue the OP had in this thread was not choosing the normal vector consistently - you can't have it pointing one way for one plane and a different way for another plane and expect to get consistent answers for the displacement :smile:

Reply 13

ghostwalker

Original post by Gome44

which tells you which side of the plane the point is on?

Yes, but you need a reference point. You can tell whether two points are on the same side or on opposite sides.

Given that ax+by+cy+d=0 is the same equation as -ax-by-cy-d=0: Plug those into your formula without the modulus signs and you'll get different results.

Reply 14

WhiteGroupMaths

I wrote this a while ago about understanding the implications of the scalar product form of vector plane equations, hope it helps.

http://www.whitegroupmaths.com/2012/08/understanding-matters-7.html

Peace.