Results 1 to 10 of 10

Math Help - Vectors Application

  1. #1
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1

    Vectors Application

    Given that a = 6i + 3j and b = 4i + tj, find the value of the real number t for which b is parallel to a.

    Note: all pronumerals are vectors.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by classicstrings View Post
    Given that a = 6i + 3j and b = 4i + tj, find the value of the real number t for which b is parallel to a.

    Note: all pronumerals are vectors.
    correct me if I'm wrong, but I think for vectors to be parrallel, they must have the same ratio of numbers...

    such as if: c=xi+yj and d=ai+bj and d is parrallel to c, then x/a=y/b


    if this is true than to solve for t you get: 6/4=3/t

    thus: 12=6t

    then: t=2

    but I haven't learned vectors, so it's your call.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by classicstrings View Post
    Given that a = 6i + 3j and b = 4i + tj, find the value of the real number t for which b is parallel to a.

    Note: all pronumerals are vectors.
    Quick is correct, but there is a more general way to do this. Two vectors are parallel if they are proportional, ie. b = ka where k is a scalar constant. So we need constants k and t such that:
    4 = 6k
    t = 3k
    (These are gotten by comparing coefficients of unit vectors i and j.)

    So from the first we get k = 4/6 = 2/3. From the second then we get that
    t = 3*2/3 = 2.

    (The reason I'm being picky about the method is that Quick's method is difficult to apply when the vectors are three or more dimensional.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1
    Quote Originally Posted by topsquark View Post
    Quick is correct, but there is a more general way to do this. Two vectors are parallel if they are proportional, ie. b = ka where k is a scalar constant. So we need constants k and t such that:
    4 = 6k
    t = 3k
    (These are gotten by comparing coefficients of unit vectors i and j.)

    So from the first we get k = 4/6 = 2/3. From the second then we get that
    t = 3*2/3 = 2.

    (The reason I'm being picky about the method is that Quick's method is difficult to apply when the vectors are three or more dimensional.)

    -Dan
    Thanks Dan! That was a rather easy question now i understand. How would I tackle this one?

    A tangent is drawn from P to a circle with centre O. It meets the circle at T. If PT = 19.8 and PM = 13.2, where M is the closest point on the circle to P, find the length of the radius.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by classicstrings View Post
    Thanks Dan! That was a rather easy question now i understand. How would I tackle this one?

    A tangent is drawn from P to a circle with centre O. It meets the circle at T. If PT = 19.8 and PM = 13.2, where M is the closest point on the circle to P, find the length of the radius.
    (Since this was a new question, it probably should go in a new thread.)

    I can't do the sketch as my computer died last week and given the trouble I was having with my scanner's software anyway, I've decided not to try to ressurect it.

    However, if you sketch the situation you should be able to verify what I'm saying.

    Sketch a line PT tangent to a circle. Now draw a radius out to point T. These two lines ALWAYS meet at a right angle.

    Now sketch the line PM. Since M is the closest point on the circle to P if we extend this line it will pass through the origin.

    What this means is that OPT is a right triangle, with legs of length r and 19.8, and a hypotenuse of length (r + 13.2).

    Thus
    r^2 + 19.8^2 = (r + 13.2)^2

    Now you can solve for r.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by classicstrings View Post
    Thanks Dan! That was a rather easy question now i understand. How would I tackle this one?

    A tangent is drawn from P to a circle with centre O. It meets the circle at T. If PT = 19.8 and PM = 13.2, where M is the closest point on the circle to P, find the length of the radius.
    From the secant-secant theorem (and if you can find an on-line reference
    for that I would like to know of it - thanks) we have:

    PT^2=PM*(PM+2r),

    or:

    19.8^2=13.2*(13.2+2r)

    so after a bit of jiggery-pokery:

    r=8.25.

    RonL

    secant-secant theorem:
    let AB and CD be two chords to a circle which meet at a point P
    external to the circle; then PA.PB=PC.PD
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1
    Quote Originally Posted by topsquark View Post
    (Since this was a new question, it probably should go in a new thread.)

    I can't do the sketch as my computer died last week and given the trouble I was having with my scanner's software anyway, I've decided not to try to ressurect it.

    However, if you sketch the situation you should be able to verify what I'm saying.

    Sketch a line PT tangent to a circle. Now draw a radius out to point T. These two lines ALWAYS meet at a right angle.

    Now sketch the line PM. Since M is the closest point on the circle to P if we extend this line it will pass through the origin.

    What this means is that OPT is a right triangle, with legs of length r and 19.8, and a hypotenuse of length (r + 13.2).

    Thus
    r^2 + 19.8^2 = (r + 13.2)^2

    Now you can solve for r.

    -Dan
    Yes I drew a diagram and it all works out perfectly, thanks again!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    From the secant-secant theorem (and if you can find an on-line reference
    for that I would like to know of it - thanks) we have:

    PT^2=PM*(PM+2r),

    or:

    19.8^2=13.2*(13.2+2r)

    so after a bit of jiggery-pokery:

    r=8.25.

    RonL

    secant-secant theorem:
    let AB and CD be two chords to a circle which meet at a point P
    external to the circle; then PA.PB=PC.PD
    Maybe this?
    Power of a point - Wikipedia, the free encyclopedia
    Power of a Point Theorem
    Circle Power -- from Wolfram MathWorld
    EDIT: That's a nice theorem! I think I will jot it down.
    Last edited by classicstrings; September 23rd 2006 at 05:41 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by classicstrings View Post
    I looked at them when I was writting my post, but I though they were
    not really stisfactory

    RonL
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1
    I'll have a dig around my sheets... i vaguely remember something ....

    EDIT: Maybe if you search "limiting case" "intersecting chords of a circle"

    http://www.nos.org/Secmathcour/eng/c...ting%20case%22

    There is something on there....page 3 20.5 as well as many nice examples

    See 20.8 as well. Actually it coves pretty much everything ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vectors Application 14
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 4th 2009, 06:12 PM
  2. Vectors Application 2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 3rd 2009, 09:26 AM
  3. Vectors Application 19
    Posted in the Calculus Forum
    Replies: 11
    Last Post: September 3rd 2009, 08:48 AM
  4. Vectors Application 3
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 31st 2009, 06:58 AM
  5. Vectors Application
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 30th 2009, 05:17 PM

Search Tags


/mathhelpforum @mathhelpforum