Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    I know the answer to the first part of the question, but I don't know why, and I don't get the second part. Can anyone help?

    A variable point P has position vector r

    a) For a non-zero constnat vector c, describe the locus of point P in the case:

    c cross product r = 0

    b) In the case that P lies on the line with Cartesian Equations

    (x-1)/2 = (y+1)/3 = (z-5)/4

    find the equation of the line in the form

    (r-a) cross product b = 0
    Offline

    2
    ReputationRep:
    a) The line through O parallel to c.

    b) The equation (r - a) x b = 0 represents the line through a parallel to b. The points (1, -1, 5) and (3, 2, 9) are on the line (x-1)/2 = (y+1)/3 = (z-5)/4. Therefore (3, 2, 9) - (1, -1, 5) = (2, 3, 4) is parallel to that line. So we can take

    a = (1, -1, 5),
    b = (2, 3, 4).
    Offline

    0
    ReputationRep:
    (Original post by Mat666)
    I know the answer to the first part of the question, but I don't know why, and I don't get the second part. Can anyone help?

    A variable point P has position vector r

    a) For a non-zero constnat vector c, describe the locus of point P in the case:

    c cross product r = 0

    b) In the case that P lies on the line with Cartesian Equations

    (x-1)/2 = (y+1)/3 = (z-5)/4

    find the equation of the line in the form

    (r-a) cross product b = 0
    For part, a), if you take the cross product, the magnitude of it is found by:
    |r||c|sin T

    Thus, if the cross product is 0 and c is non-zero, then either r is 0 (in which case the locus is just the origin), or sin T=0, in which case the locus is a line through the origin in the direction of c (which also includes the point in the origin), so the solution is a line through the origin in the direction of c.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thanks: just a little clearing up. I understand how you got the points on the line, but why take one away from the other? I'm being really stupid today and its too hot, I usually get P6 vectors...
    Offline

    2
    ReputationRep:
    (Original post by Mat666)
    Thanks: just a little clearing up. I understand how you got the points on the line, but why take one away from the other? I'm being really stupid today and its too hot, I usually get P6 vectors...
    Consider the line L in 2d that goes through the points (0, 1) and (1, 0). By drawing a picture you can see that L is parallel to the vector (1, 0) - (0, 1) = (1, -1). It's the same in 3d (but more difficult to draw).
    Offline

    0
    ReputationRep:
    (Original post by Mat666)
    Thanks: just a little clearing up. I understand how you got the points on the line, but why take one away from the other? I'm being really stupid today and its too hot, I usually get P6 vectors...
    All you need is the position vector of a point on the line (a) and a point on the line (b). You can find a by finding any point that satisfies the equation of the line. To get b, all you need is two such points that satisfy the equation of the line then find the vector which gets you from one to the other, as this is in the direction of the line. If r is on the line, then r-a will be parallel to b.
    • Thread Starter
    Offline

    0
    ReputationRep:
    No, no, I get you now, thanks. They always ask dumb questions on P6, I just prey for the straight forward 'find the inverse of this matrix' questions...
    Offline

    0
    ReputationRep:
    (Original post by Mat666)
    No, no, I get you now, thanks. They always ask dumb questions on P6, I just prey for the straight forward 'find the inverse of this matrix' questions...
    The good news is that the standardisation means you don't have to get as many real marks as usual to secure the same grade as the paper is harder (which is why I ended up with a better mark on P6 than D1 last year. )
 
 
 
Turn on thread page Beta
Updated: June 8, 2004

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?
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.