Results 1 to 7 of 7

Math Help - Equation of a line

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    290

    Equation of a line

    I don't get the explanation.

    Given that line k passes through point A(4,-5,3) and has direction vector (-2,3,0), the equation of line k can be expressed as:

    \frac {x-4}{-2}=\frac {y+5}{3}=\frac {z-3}{0} (1)

    If only the denominator is 0, then equation 1 is meaningless. (What does that mean)

    If when the denominator is 0, we assume that the numerator is also 0 (why?) and the fraction can be any value (why?), then equation 1 can be expressed as:

    \frac {x-4}{-2}=\frac {y+5}{3} (2)

    z=3 (how do you get that?)

    Equations 1 and 2 mean the same thing. Equation 2 is usually used.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    The usual way to write this is \frac{x-4}{-2}=\frac{y+5}{3};~z=3.
    The z-value is constant. There is no change in the z-direction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,714
    Thanks
    1472
    You can also write that as "parametric equations":
    x= 4- 2t, y= -5+ 3t, z= 3+ 0t= 3

    Solving the first two equations for t, t= (4- x)/2 and t= (5+y)/3. Set those equal: (4-x)/2= (5+y)/2. Since there is no t in the z equation, you cannot solve for t- but you still have z= 3.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    290
    Do you mind answering the questions I wrote in my first post? There is a reason I wrote them.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    Quote Originally Posted by chengbin View Post
    Do you mind answering the questions I wrote in my first post? There is a reason I wrote them.
    What questions do you think that the two of us did not answer?
    Because of this question, I wonder if you even understand the question.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,714
    Thanks
    1472
    Ahh!
    If when the denominator is 0, we assume that the numerator is also 0 (why?) and the fraction can be any value (why?)
    z=3 (how do you get that?)
    \frac{a}{0}= x is the same as a= 0(x). But 0 times any number is 0 so that only makes sense if a= 0- that is, if "the numerator is also 0". but in that case 0= 0(x) for any x. That is why "the fraction can be any value". So for \frac{z- 3}{0} to make any sense, the numerator must be 0: z- 3= 0 so z= 3.

    That's abusing the notation slightly. In fact, you can't divide by 0. All of this should be done in terms of limits to be valid.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jan 2009
    Posts
    290
    Quote Originally Posted by HallsofIvy View Post
    Ahh!

    \frac{a}{0}= x is the same as a= 0(x). But 0 times any number is 0 so that only makes sense if a= 0- that is, if "the numerator is also 0". but in that case 0= 0(x) for any x. That is why "the fraction can be any value". So for \frac{z- 3}{0} to make any sense, the numerator must be 0: z- 3= 0 so z= 3.

    That's abusing the notation slightly. In fact, you can't divide by 0. All of this should be done in terms of limits to be valid.
    Thanks a lot for clearing this up. Just what I needed to understand this.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 18th 2011, 10:36 AM
  2. Replies: 9
    Last Post: July 31st 2009, 05:39 AM
  3. Replies: 11
    Last Post: June 2nd 2009, 06:08 PM
  4. Replies: 5
    Last Post: October 13th 2008, 10:16 AM
  5. equation of a line perpendicular to the line defined
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 6th 2007, 09:02 AM

Search Tags


/mathhelpforum @mathhelpforum