The Student Room Group

Vectors

17. Sketch the plane r Β· n = 1, where n is a unit vector parallel to i + j + k. If the plane
represents a perfectly reflecting surface (a mirror), and a thin light beam is travelling
towards it along the straight line
r = i + ks,
find the point at which the beam hits the mirror. What is the equation representing
the path of the reflected ray?

What I've done so far

Spoiler



Please don't post a solution, just hints please :smile:

Scroll to see replies

Reply 1
Original post by Rock_Set
17. Sketch the plane r Β· n = 1, where n is a unit vector parallel to i + j + k. If the plane
represents a perfectly reflecting surface (a mirror), and a thin light beam is travelling
towards it along the straight line
r = i + ks,
find the point at which the beam hits the mirror. What is the equation representing
the path of the reflected ray?

What I've done so far

Spoiler



Please don't post a solution, just hints please :smile:


trying to check your workings ...

is the line r = i + sk ?
Reply 2
Original post by TeeEm
trying to check your workings ...

is the line r = i + sk ?



Yes :smile:
Reply 3
Original post by Rock_Set
Yes :smile:


your workings are all correct up to there (assuming the sketch is correct)
Reply 4
Original post by TeeEm
your workings are all correct up to there (assuming the sketch is correct)


My sketch is 3 lines connecting root 3 on the x, y and z axis. As shown below, but root 3

http://4.bp.blogspot.com/_f3d3llNlZKQ/TFspNGqS8VI/AAAAAAAACgU/HJltdORX814/s400/E.png

I'll try and continue then if I'm correct so far and see what I get, thank you :smile:

I'll post back when I've done
Reply 5
Original post by TeeEm
your workings are all correct up to there (assuming the sketch is correct)


do you need help with rest?

or just checking work?
Reply 6
Original post by Rock_Set
My sketch is 3 lines connecting root 3 on the x, y and z axis. As shown below, but root 3

http://4.bp.blogspot.com/_f3d3llNlZKQ/TFspNGqS8VI/AAAAAAAACgU/HJltdORX814/s400/E.png

I'll try and continue then if I'm correct so far and see what I get, thank you :smile:

I'll post back when I've done


all good
Reply 7
Original post by TeeEm
all good


I've had a go and I keep getting lost and having to start again. Could you give me a hint on how to proceed please? Thanks :smile:
Reply 8
be back in ten
Original post by Rock_Set
I've had a go and I keep getting lost and having to start again. Could you give me a hint on how to proceed please? Thanks :smile:
You need to find the direction of the reflected ray.

It will help to draw a diagram: if d is the incoming direction (i.e. (0,0,1)) and e the outgoing direction, the key thing is to work out e-d.

Hint: what vector is e-d parallel to?
Reply 10
Original post by DFranklin
You need to find the direction of the reflected ray.

It will help to draw a diagram: if d is the incoming direction (i.e. (0,0,1)) and e the outgoing direction, the key thing is to work out e-d.

Hint: what vector is e-d parallel to?


The normal?
Reply 11
Original post by TeeEm
be back in ten


sorry phone

now the problem can be done with cross products but it is easier to use simple geometry.

pick point A on your line by inspection

work perpendicular line from that point to the plane.

find intersection with plane, point M

M is midpoint of AB, B being reflection of A "below plane.

then you have the original intersection found in part (a) and point B to find what you need.


Draw cross-sectional diagram


hope this helps
Reply 12
Original post by DFranklin
You need to find the direction of the reflected ray.

It will help to draw a diagram: if d is the incoming direction (i.e. (0,0,1)) and e the outgoing direction, the key thing is to work out e-d.

Hint: what vector is e-d parallel to?


Original post by TeeEm
be back in ten


I let my reflected vector = r2 = ai + bj + ck

r - r2 = (1-a)i + (-b)j + (root(3) - 1 - c)k and is parallel to n

(r - r2) cross n = 0

Used these equations to get that r2 = ai + (a-1)j + (a - 2 + root(3))k

(r - r2) dot n = 0

I then said r dot n = r2 dot n

Thus r2 dot n = 1

I did this and got 3a = 3, so a = 1

I subbed this back in and got r2 = i + (-1+root(3))k

I've read the geometric method and that seems much easier, I think I understand what you're doing there.
(edited 9 years ago)
Note that there is a distinction between the direction of the reflected ray and the position of a point on the reflected ray. As far as I can see, your calculations are based on position, but when calculating reflections (or refractions, if you ever need to do that), you want to work with the direction, not position. (And then at the end when you want an equation for the reflected ray, use the direction and combine with a point you know is on the reflected ray, such as the point where the original ray intersects the plane).

You should not need to use a cross product (and if I saw a cross product in someone's working for this I would immediately suspect they had done something wrong).
(edited 9 years ago)
Reply 14
Original post by DFranklin
Note that there is a distinction between the direction of the reflected ray and the position of a point on the reflected ray. As far as I can see, your calculations are based on position, but when calculating reflections (or refractions, if you ever need to do that), you want to work with the direction, not position. (And then at the end when you want an equation for the reflected ray, use the direction and combine with a point you know is on the reflected ray, such as the point where the original ray intersects the plane).

You should not need to use a cross product (and if I saw a cross product in someone's working for this I would immediately suspect they had done something wrong).


Oh... That's annoying, I thought I'd managed it. I'll have another go and see what happens :smile:
Reply 15
Original post by DFranklin
Note that there is a distinction between the direction of the reflected ray and the position of a point on the reflected ray. As far as I can see, your calculations are based on position, but when calculating reflections (or refractions, if you ever need to do that), you want to work with the direction, not position. (And then at the end when you want an equation for the reflected ray, use the direction and combine with a point you know is on the reflected ray, such as the point where the original ray intersects the plane).

You should not need to use a cross product (and if I saw a cross product in someone's working for this I would immediately suspect they had done something wrong).


Hey, I've been trying for hours now and so far I've made no progress. Could you provide some more hints please? I have a+b+c=-1, for the direction vector of the reflected light ray.
Reply 16
Original post by Rock_Set
Hey, I've been trying for hours now and so far I've made no progress. Could you provide some more hints please? I have a+b+c=-1, for the direction vector of the reflected light ray.


probably it is a better idea to wait until DFranklin comes back as you are trying to follow his advice. If you are not successful look again at post 12 of your thread.
Reply 17
Original post by TeeEm
probably it is a better idea to wait until DFranklin comes back as you are trying to follow his advice. If you are not successful look again at post 12 of your thread.


I think I've done it using your method. I got

r(reflected) = (i + (root3 - 1)k) + t(-2i+k)
Reply 18
Original post by Rock_Set
I think I've done it using your method. I got

r(reflected) = (i + (root3 - 1)k) + t(-2i+k)


I cannot check your answer right now ...

I posted hours ago but you did not see it ...

all good now.
Reply 19
Original post by TeeEm
I cannot check your answer right now ...

I posted hours ago but you did not see it ...

all good now.


Yes sorry.

I now have two answers, not sure which is correct

r(reflected) = (i + (root3 - 1)k) + t(-2i+k)

or

r(reflected) = (i + (root3 - 1)k) + t(-2i-2j+k)

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