Results 1 to 3 of 3

Math Help - Calc III

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    25

    Calc III

    Given the lines : L1: x = -1 - 2t, y = -6t, z = 8t
    L2: x = 3 + t, y = 3t, z = 5 - 4t

    (a) Show that the lines are parallel.
    (b) Find the distance between the two lines.
    (c) Find the equation of the plane containing the two lines.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Aug 2009
    Posts
    143
    Quote Originally Posted by tdat1979 View Post
    Given the lines : L1: x = -1 - 2t, y = -6t, z = 8t
    L2: x = 3 + t, y = 3t, z = 5 - 4t

    (a) Show that the lines are parallel.
    (b) Find the distance between the two lines.
    (c) Find the equation of the plane containing the two lines.
    (a)  \vec{n}_1=\left[\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right]=[-2, -6, 8] \ \ \<br />
\vec{n}_2=[ 1, 3, -4] \ \ \vec{n}=\frac{[1, 3, -4]}{\sqrt{26}}

    (b) Pick a point on L1, say (-1, 0, 0) and another point on L2, say (3, 0, 5). The vector is \vec{r}=[4, 0, 5] \ \ \ d=|\vec{r} \times \vec{n}|

    (c) The normal vector of the plane is \vec{r} \times \vec{n}
    Last edited by mr fantastic; September 18th 2009 at 09:04 AM. Reason: Restored original post
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    a.

    For L1 v1 = -2 i - 6j +8k

    L2 v2 = i + 3 j - 4k

    since v1 = -2v2 they are parallel

    b. consider the vector v from (3,0,5) on L2 to (-1,0,0) on L1

    Then the distance between the 2 lines is the magnitude of v - projection of v on v2

    See attachment

    v = 4 i + 5k

    projv on v2 = (4 i + 5k)*(i + 3j -4k)/26 *(i +3j -4k)

    = (-16/26)(i +3j -4k)


    v - projv on v2 = (4+16/26)i +48/26j + (5-64/26)k

    |v- projv on v1 | = 5.58


    c. Use 2 points on L1 (-1,0,0) and (1,6,-8)

    one point on L2 (3,0,5)

    Let u be the vector from (-1,0,0) to (3,0,5) and v be the vector

    from (1,6,-8) to (3,0,5)

    Then the normal to the desired plane is N = u x v = ai +bj+ck

    using any of the 3 points use a(x-x0) +b(y-y0) +c(z-z0) = 0

    to generate the eqn of the plane
    Attached Thumbnails Attached Thumbnails Calc III-vector.jpg  
    Last edited by Calculus26; September 8th 2009 at 06:28 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calc 1 review for Calc 2
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 28th 2011, 04:47 AM
  2. A few Pre-Calc problems in my AP Calc summer packet
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 30th 2010, 04:40 PM
  3. Replies: 1
    Last Post: January 13th 2010, 12:57 PM
  4. What am I doing? Calc, Pre-calc, Trig, etc...
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 23rd 2008, 09:51 AM
  5. Adv Calc lim prf
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 6th 2007, 10:25 AM

Search Tags


/mathhelpforum @mathhelpforum