Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    line equation, L: (x, y, z) = (1, 1, -3) + t(1, 2, 3)
    plane equation, P: -10z - y + 4z = 0

    find distance between these 2
    Offline

    14
    ReputationRep:
    (Original post by MalaysianDude)
    line equation, L: (x, y, z) = (1, 1, -3) + t(1, 2, 3)
    plane equation, P: -10z - y + 4z = 0

    find distance between these 2
    hmm i hope this helps

    first change the equation of the plane
    r.(-10i - j + 4k) = a.n = (i +2j + 3k).(-10i -j +4k)
    which implies r.(-10i - j + 4k) = -10 -2 + 12 = 0
    hmm tricky
    iv confused myself
    see what u can do from here
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by kriztinae)
    hmm i hope this helps

    first change the equation of the plane
    r.(-10i - j + 4k) = a.n = (i +2j + 3k).(-10i -j +4k)
    which implies r.(-10i - j + 4k) = -10 -2 + 12 = 0
    hmm tricky
    iv confused myself
    see what u can do from here
    nopes
    that doesnt help at all
    i got stuck there as well
    so can someone help me out here?
    Offline

    14
    ReputationRep:
    (Original post by MalaysianDude)
    nopes
    that doesnt help at all
    i got stuck there as well
    so can someone help me out here?
    check the question again, sometimes i copy things out wrong
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by kriztinae)
    check the question again, sometimes i copy things out wrong
    no
    its not wrong
    its the same question
    and its very small
    they just want you to find the distance
    thats all they gave
    Offline

    14
    ReputationRep:
    (Original post by MalaysianDude)
    no
    its not wrong
    its the same question
    and its very small
    they just want you to find the distance
    thats all they gave
    well do u have the answer?
    maybe the distance is zero coz the line intersects with the plane?
    Offline

    0
    ReputationRep:
    As they are asking for the distance between the two, they must be parallel (as otherwise they would intersect).

    Therefore, just find the distance between the point on the line (1,1,-3) and the line using
    Code:
    ae+bf+cg-d
    d= -----------------
       sqrt(a^2+b^2+c^2)
    where (e,f,g) is your point and ax+by+cz=d is your plane.



    Of course it could be a trick question as kriztinae suggests, whereby they actually intersect giving d=0.
    Offline

    2
    ReputationRep:
    My other post was wrong - I was using the wrong angle - here's the right way.

    The line and the plane don't intersect (fairly obvious from the question but you can check by subbing the general point on the line into the equation of the plane), so the line will be parallel to the plane and therefore a fixed distance from it. The shortest distance will be perpendicular to the plane.

    If you then draw a sketch diagram (attached), you can see that h = |a|cosx where a is just any vector joining a point on the plane and a point on the line. The dot product of a.n where n is the normal vector to the plane is a.n = |a||n|cosx so |a|cosx = h = (a.n)/|n| = a.m where m is unit vector perpendicular to the plane.

    The plane passes through the origin so you can take the vector a to be (1,1,-3) - the vector from O to the point on the line with t=0. m = (-10,-1,4)/(sqrt127) so h = (-10 - 1 - 12)/(sqrt127) = -23/(sqrt127) As h is a distance it should obviously be positive so distance of line from plane is 23/(sqrt127) which is the same answer as zaphods formula gives. My method is basically the working behind this formula.
    Attached Images
     
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Bezza)
    My other post was wrong - I was using the wrong angle - here's the right way.

    The line and the plane don't intersect (fairly obvious from the question but you can check by subbing the general point on the line into the equation of the plane), so the line will be parallel to the plane and therefore a fixed distance from it. The shortest distance will be perpendicular to the plane.

    If you then draw a sketch diagram (attached), you can see that h = |a|cosx where a is just any vector joining a point on the plane and a point on the line. The dot product of a.n where n is the normal vector to the plane is a.n = |a||n|cosx so |a|cosx = h = (a.n)/|n| = a.m where m is unit vector perpendicular to the plane.

    The plane passes through the origin so you can take the vector a to be (1,1,-3) - the vector from O to the point on the line with t=0. m = (-10,-1,4)/(sqrt127) so h = (-10 - 1 - 12)/(sqrt127) = -23/(sqrt127) As h is a distance it should obviously be positive so distance of line from plane is 23/(sqrt127) which is the same answer as zaphods formula gives. My method is basically the working behind this formula.
    thats the right answer man
    anyways,did you use the formula given by Zaphod?
    Offline

    0
    ReputationRep:
    For OCR at least, the formula I quoted is given (albeit in a form I can't easily reproduce here). So you'd have to check whether or not it would be given on your syllabus as to whether it's useable like that or not.
    Offline

    2
    ReputationRep:
    (Original post by MalaysianDude)
    thats the right answer man
    anyways,did you use the formula given by Zaphod?
    I didn't use the formula, I just thought about it and drew a diagram. I checked in the formula though and realised I'd done it wrong the first time - I'd used x as the angle between the vector a and the plane, rather than a and the normal vector to the plane.

    The formula is given in the edexcel formula book too.
    Offline

    1
    ReputationRep:
    the easiest way to solve vector problems like this is to draw a diagram.

    then you can see what you have and hence whether to use the cross or dot product
 
 
 
Turn on thread page Beta
Updated: June 14, 2004
Poll
Are you going to a festival?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.