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Thread: Need help with symmetric equation

  1. #1
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    Need help with symmetric equation

    Write a symmetric equation and parametric equation for
    a) Line passes through the point A (4,1) with direction vector m = [-3, 1]

    b) passes through the point R(-6, 2) and S(4, -2)

    2) The equation of a line is 2x-3y+12=0. Write equation for line in parametric form

    Thanks in advance

    *plz also explain them to me
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  2. #2
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    b) passes through the point R(-6, 2) and S(4, -2)


    First, we need a vector, v that will be parallel to the line. We thus let v be the vector that starts at point S and ends at point R.

    v = <-6-4, 2-(-2)> = <-10, 4>.

    Now that we have our direction vector v, all we need to do is choose of our two given points, R or S. Let's use R.

    The equation of our line r = <-6, 2> + t<-10, 4> = <-6-10t, 2+4t>.

    In parametric form:
    $\displaystyle x = -6 - 10t $
    $\displaystyle y = 2 + 4t $

    In symmetric form:
    $\displaystyle \frac{x+6}{-10} = \frac{y-2}{4} $.
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  3. #3
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    Thanks for the explanation

    now need part a and 2
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  4. #4
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    a) Line passes through the point A (4,1) with direction vector m = [-3, 1]


    We have a direction vector, m and a point to work with. This should not be difficult.

    r = <4,1> + t<-3,1> = <4-3t, 1+t>

    parametric form:
    $\displaystyle x = 4 - 3t $
    $\displaystyle y = 1 + t $

    symmetric form:
    $\displaystyle \frac{x-4}{-3} = \frac{y-1}{1} $
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  5. #5
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    Thanks again

    now need help with last question
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  6. #6
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    2) The equation of a line is 2x-3y+12=0. Write equation for line in parametric form.

    What we are given here is an implicit equation of a line. To convert from an implicit representation to a parametric representation, follow my steps below:

    We let $\displaystyle \vec r $ be the position vector, $\displaystyle <x, y> $ and let $\displaystyle \vec n $ be the vector of coefficients, $\displaystyle <2, -3> $.

    That is,
    $\displaystyle \vec r = <x, y> $
    $\displaystyle \vec n = <2, -3> $

    Now, we can rewrite 2x-3y+12=0 in terms of $\displaystyle \vec r $ and $\displaystyle \vec n $.

    $\displaystyle \vec n \cdot \vec r = -12 $

    To write this in vector form, we must now do two things. A) we must find a point, $\displaystyle \vec{r_{0}} $ on the line and B) find a vector $\displaystyle \vec v $ parallel to the line.

    To find any point on 2x-3y+12=0, we can simply just use the x-intercept. That is, we simultaneously solve for 2x-3y+12=0 and x=0.

    $\displaystyle 2x - 3y + 12 = 0 $

    $\displaystyle x = 0 $

    $\displaystyle -3y + 12 = 0 $

    $\displaystyle -3y = -12 $

    $\displaystyle 3y = 12 $

    $\displaystyle y = \frac{12}{3} $ = 4

    Therefore, we let point $\displaystyle \vec{r_{0}} = <0, 4> $.

    Now, for part B that I mentioned above, find a vector $\displaystyle \vec v $ parallel to the line.

    $\displaystyle \vec n = <2, -3> $

    $\displaystyle \vec v = \vec{n_{\perp}} = <-3, -2> $

    The general parametric equation in vector form is $\displaystyle \vec{r}(t) = \vec{r_{0}} + t \vec v $.

    Thus, $\displaystyle \vec{r}(t) = <0, 4> + t<-3, -2> $.

    Finally, in parametric non-vector form, we have:

    x = -3t
    y = 4 - 2t

    Good luck!
    -Andy
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  7. #7
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    And to double-check my solution for question 2, let's substitute x=-3t and y=4-2t in 2x-3y+12 = 0:

    $\displaystyle 2x-3y+12 = 0 $

    $\displaystyle 2(-3t) - 3(4-2t) + 12 = 0 $

    $\displaystyle -6t -12 + 6t + 12 = 0 $

    And we know we are correct...

    -Andy
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  8. #8
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    Thanks Andy
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  9. #9
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    Vectors symmetric equations

    This question is givin me trouble big tim

    Que/

    Find symmetric equations for the line

    x=2+3t, y=4-4t, z=-1+6t


    If anyone can solve this, itll be very greatful..
    thanx
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  10. #10
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    Quote Originally Posted by K-D View Post
    This question is givin me trouble big tim

    Que/

    Find symmetric equations for the line

    x=2+3t, y=4-4t, z=-1+6t


    If anyone can solve this, itll be very greatful..
    thanx
    As abender has shown in the first reply you only have to solve each equation for t and set them equal:

    $\displaystyle x=2+3t~\implies~\boxed{t=\dfrac{x-2}3}$
    $\displaystyle y=4-4t~\implies~\boxed{t=\dfrac{y-4}4}$
    $\displaystyle z=-1+6t~\implies~\boxed{t=\dfrac{z+1}6}$

    Thus the line in symmetric form is:

    $\displaystyle \dfrac{x-2}3 = \dfrac{y-4}4 = \dfrac{z+1}6$
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