Results 1 to 5 of 5

Math Help - Vector help

  1. #1
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288

    Vector help

    GCE Maths
    With respect to an origin O, the pts. A and B have the following position vectors

    OA=2i+2j+k OB=i+4j+3k

    The line has vector equation
    r=4i-2j+2k+s(i+2j+k)

    I. Prove that the line l does not intersect the line through A and B

    II. Find the equation of the plane containing l and the point A giving your answer in the form ax+by+cz=d
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    180

    Some explaining please

    Quote Originally Posted by I-Think View Post
    GCE Maths
    With respect to an origin O, the pts. A and B have the following position vectors

    OA=2i+2j+k OB=i+4j+3k

    The line has vector equation
    r=4i-2j+2k+s(i+2j+k) what is s?
    is this "line" a sum of the vectors OA;OB? if so is the point of the exercise to find s?

    I. Prove that the line l does not intersect the line through A and B
    If a and b are vectors it does not matter where on the cordinate system you place them. Not sure what you mean here...?

    II. Find the equation of the plane containing l \Rightarrow what is this? and the point A giving your answer in the form ax+by+cz=d
    \Rightarrow is that the sum of vectors OA;OB?
    Regards H (sorry but really difficult to understand without a picture)..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello I-Think
    Quote Originally Posted by I-Think View Post
    GCE Maths
    With respect to an origin O, the pts. A and B have the following position vectors

    OA=2i+2j+k OB=i+4j+3k

    The line has vector equation
    r=4i-2j+2k+s(i+2j+k)

    I. Prove that the line l does not intersect the line through A and B

    II. Find the equation of the plane containing l and the point A giving your answer in the form ax+by+cz=d
    First we need to find the vector equation of the line through A and B. This can be written as
    \vec r = \vec{OA}+t\vec{AB} ...(1)
    where t is a scalar parameter. Do you understand this bit? It simply says:
    \underbrace{\vec r=}_{\text{You can move from O to any point on the line ... }} \underbrace{\vec{OA}}_{\text{ by a movement to A ...}}\underbrace{+}_{\text{ followed by ...}}\underbrace{t\vec {AB}}_{\text{ a movement in the direction of AB}}
    Now
    \vec{OA} = 2\vec i + 2\vec j + \vec k
    and
    \vec{AB} = \vec{OB} - \vec{OA}
    = -\vec i +2 \vec j +2\vec k
    So, from (1), the equation of the line through AB is:
    \vec r = 2\vec i + 2\vec j + \vec k+t(-\vec i +2 \vec j +2\vec k)
    Next, we need to prove that this line doesn't intersect line l. We do this by showing that there are no values of s and t that will bring us to the same point on each of the lines. So, suppose that the lines meet. Then, for some value of s and of t:
    2\vec i + 2\vec j + \vec k+t(-\vec i +2 \vec j +2\vec k)= 4\vec i-2\vec j+2\vec k+s(\vec i+2\vec j+\vec k)
    Comparing the coefficients of \vec i and \vec j, we get:
    2-t = 4+s
    and
    2+2t = -2+2s
    which gives:
    s = 0,\;t=-2
    If we now compare the coefficients of \vec k:
    On the LHS: 1-4 = -3

    On the RHS: 2 + 0 = 2
    Since these are not equal, there are no values of s and t that satisfy the equation, and hence the lines do not intersect.

    Part II

    We have a choice of two methods here. The first uses simple algebra; the second, rather more sophisticated, uses a vector and scalar product.

    For the first method, we find a couple of points on the line l by taking two values of s, and then solve a set of simultaneous equations.

    So, with s = 0 and s = 1, we get the points (4, -2, 2) and (5, 0, 3) lying on l, and therefore also in the plane. And the point A also lies in the plane.

    So, if the plane is ax+by+cz=1 (which is simply the equation ax+by+cz=d re-written when divided throughout by d, using new values of a, b, c), we get:
    \left \{\begin{array}{l}<br />
4a-2b+2c=1\\<br />
5a+0b+3c=1\\<br />
2a+2b+c=1\\<br />
\end{array}\right.
    This gives:
    \left \{\begin{array}{l}<br />
a=1\\<br />
b=\tfrac16\\<br />
c=-\tfrac43\\<br />
\end{array}\right.
    and, when we get rid of fractions, gives the plane:
    6x+y-8z=6
    If you want to see an explanation of the vector method, look at this page.

    Grandad
    Last edited by Grandad; February 9th 2010 at 06:59 AM. Reason: I missed out a letter z
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    180

    Well

    Wow!!
    Grandad is the master
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288
    Grandad, the day I learn to teach as well as you is the day I will be a better person
    Last edited by I-Think; February 21st 2010 at 02:29 AM. Reason: Spelling errors, my english teacher will be appalled
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: January 11th 2011, 10:17 AM
  2. Replies: 2
    Last Post: January 27th 2010, 08:23 PM
  3. Replies: 11
    Last Post: December 23rd 2009, 01:30 AM
  4. Replies: 4
    Last Post: November 5th 2009, 04:03 PM
  5. Replies: 2
    Last Post: October 5th 2009, 03:25 PM

Search Tags


/mathhelpforum @mathhelpforum