Equation of line: find $p + q$












0














The question: The following two lines intersect, forming an angle of $60°$:
$$ frac{x-1}a = frac{y-2}{a+1} = frac{z-1}{a-1} \
x = y ; & ; z = 1. $$

If $a = -(p/q)$ where $p$ and $q$ are coprime positive integers, what is $p + q$? Assume that $a$ is not equal to $0$, $-1$, $1$.



My solution:

Direction vector of line 1: $v_1(a,a+1,a-1)$

Direction vector of line 2: $v_2(1, 1, 0)$

Unit vector $hat a$ (arbitrary) : $left(1/sqrt3; 1/sqrt3; 1/sqrt3right)$




  1. Cross-product: $$ begin{align} v_3 = v_1 times v_2 &= left(sqrt{a^2+(a+1)^2+(a-1)^2} cdot sqrt2 cdot
    sin{60°}right) hat a \ &= left(frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}right). tag 1 end{align}$$


  2. Cross-product with coordinates: coordinate $x$ of $v_3 = (a+1)cdot0 - (a-1)cdot1 = 1-a. tag2$



From (1) and (2) => $$ begin{align} &; frac12sqrt{6a^2+4} = 1-a \
⇔ sqrt{6a^2+4} = 2-2a \
⇔ a = -4 text{ (checked)}. end{align}$$



But $a = -p/q$ (coprime positive integers). So my solution is wrong.

Anyone knows how to solve it? Please give hint or instructions! Thanks!






analytic-geometry







share|cite|improve this question
























  • Please make your question self-contained instead of linking to another site.
    – amd
    Nov 26 at 23:54










  • That would be so lòng. Please check my edited link. Thanks
    – Willy
    Nov 27 at 13:08










  • Part of the function of this site is as an archive. External links go stale, so some day no one will have any idea what your question actually was when they come to this post. Aside from that, you’re asking others to take their time to help you. It’s only fair that you take some time of your own to formulate your question properly.
    – amd
    Nov 27 at 20:04










  • Thanks @ and. Lesson learned. I edited my question.
    – Willy
    Nov 28 at 5:33










  • The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $xne y$.
    – amd
    Nov 28 at 6:14


















0














The question: The following two lines intersect, forming an angle of $60°$:
$$ frac{x-1}a = frac{y-2}{a+1} = frac{z-1}{a-1} \
x = y ; & ; z = 1. $$

If $a = -(p/q)$ where $p$ and $q$ are coprime positive integers, what is $p + q$? Assume that $a$ is not equal to $0$, $-1$, $1$.



My solution:

Direction vector of line 1: $v_1(a,a+1,a-1)$

Direction vector of line 2: $v_2(1, 1, 0)$

Unit vector $hat a$ (arbitrary) : $left(1/sqrt3; 1/sqrt3; 1/sqrt3right)$




  1. Cross-product: $$ begin{align} v_3 = v_1 times v_2 &= left(sqrt{a^2+(a+1)^2+(a-1)^2} cdot sqrt2 cdot
    sin{60°}right) hat a \ &= left(frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}right). tag 1 end{align}$$


  2. Cross-product with coordinates: coordinate $x$ of $v_3 = (a+1)cdot0 - (a-1)cdot1 = 1-a. tag2$



From (1) and (2) => $$ begin{align} &; frac12sqrt{6a^2+4} = 1-a \
⇔ sqrt{6a^2+4} = 2-2a \
⇔ a = -4 text{ (checked)}. end{align}$$



But $a = -p/q$ (coprime positive integers). So my solution is wrong.

Anyone knows how to solve it? Please give hint or instructions! Thanks!






analytic-geometry







share|cite|improve this question
























  • Please make your question self-contained instead of linking to another site.
    – amd
    Nov 26 at 23:54










  • That would be so lòng. Please check my edited link. Thanks
    – Willy
    Nov 27 at 13:08










  • Part of the function of this site is as an archive. External links go stale, so some day no one will have any idea what your question actually was when they come to this post. Aside from that, you’re asking others to take their time to help you. It’s only fair that you take some time of your own to formulate your question properly.
    – amd
    Nov 27 at 20:04










  • Thanks @ and. Lesson learned. I edited my question.
    – Willy
    Nov 28 at 5:33










  • The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $xne y$.
    – amd
    Nov 28 at 6:14
















0












0








0







The question: The following two lines intersect, forming an angle of $60°$:
$$ frac{x-1}a = frac{y-2}{a+1} = frac{z-1}{a-1} \
x = y ; & ; z = 1. $$

If $a = -(p/q)$ where $p$ and $q$ are coprime positive integers, what is $p + q$? Assume that $a$ is not equal to $0$, $-1$, $1$.



My solution:

Direction vector of line 1: $v_1(a,a+1,a-1)$

Direction vector of line 2: $v_2(1, 1, 0)$

Unit vector $hat a$ (arbitrary) : $left(1/sqrt3; 1/sqrt3; 1/sqrt3right)$




  1. Cross-product: $$ begin{align} v_3 = v_1 times v_2 &= left(sqrt{a^2+(a+1)^2+(a-1)^2} cdot sqrt2 cdot
    sin{60°}right) hat a \ &= left(frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}right). tag 1 end{align}$$


  2. Cross-product with coordinates: coordinate $x$ of $v_3 = (a+1)cdot0 - (a-1)cdot1 = 1-a. tag2$



From (1) and (2) => $$ begin{align} &; frac12sqrt{6a^2+4} = 1-a \
⇔ sqrt{6a^2+4} = 2-2a \
⇔ a = -4 text{ (checked)}. end{align}$$



But $a = -p/q$ (coprime positive integers). So my solution is wrong.

Anyone knows how to solve it? Please give hint or instructions! Thanks!






analytic-geometry







share|cite|improve this question















The question: The following two lines intersect, forming an angle of $60°$:
$$ frac{x-1}a = frac{y-2}{a+1} = frac{z-1}{a-1} \
x = y ; & ; z = 1. $$

If $a = -(p/q)$ where $p$ and $q$ are coprime positive integers, what is $p + q$? Assume that $a$ is not equal to $0$, $-1$, $1$.



My solution:

Direction vector of line 1: $v_1(a,a+1,a-1)$

Direction vector of line 2: $v_2(1, 1, 0)$

Unit vector $hat a$ (arbitrary) : $left(1/sqrt3; 1/sqrt3; 1/sqrt3right)$




  1. Cross-product: $$ begin{align} v_3 = v_1 times v_2 &= left(sqrt{a^2+(a+1)^2+(a-1)^2} cdot sqrt2 cdot
    sin{60°}right) hat a \ &= left(frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}; frac12sqrt{6a^2+4}right). tag 1 end{align}$$


  2. Cross-product with coordinates: coordinate $x$ of $v_3 = (a+1)cdot0 - (a-1)cdot1 = 1-a. tag2$



From (1) and (2) => $$ begin{align} &; frac12sqrt{6a^2+4} = 1-a \
⇔ sqrt{6a^2+4} = 2-2a \
⇔ a = -4 text{ (checked)}. end{align}$$



But $a = -p/q$ (coprime positive integers). So my solution is wrong.

Anyone knows how to solve it? Please give hint or instructions! Thanks!






analytic-geometry




analytic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 6:36









amd

29k21050




29k21050










asked Nov 26 at 10:45









Willy

414




414












  • Please make your question self-contained instead of linking to another site.
    – amd
    Nov 26 at 23:54










  • That would be so lòng. Please check my edited link. Thanks
    – Willy
    Nov 27 at 13:08










  • Part of the function of this site is as an archive. External links go stale, so some day no one will have any idea what your question actually was when they come to this post. Aside from that, you’re asking others to take their time to help you. It’s only fair that you take some time of your own to formulate your question properly.
    – amd
    Nov 27 at 20:04










  • Thanks @ and. Lesson learned. I edited my question.
    – Willy
    Nov 28 at 5:33










  • The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $xne y$.
    – amd
    Nov 28 at 6:14




















  • Please make your question self-contained instead of linking to another site.
    – amd
    Nov 26 at 23:54










  • That would be so lòng. Please check my edited link. Thanks
    – Willy
    Nov 27 at 13:08










  • Part of the function of this site is as an archive. External links go stale, so some day no one will have any idea what your question actually was when they come to this post. Aside from that, you’re asking others to take their time to help you. It’s only fair that you take some time of your own to formulate your question properly.
    – amd
    Nov 27 at 20:04










  • Thanks @ and. Lesson learned. I edited my question.
    – Willy
    Nov 28 at 5:33










  • The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $xne y$.
    – amd
    Nov 28 at 6:14


















Please make your question self-contained instead of linking to another site.
– amd
Nov 26 at 23:54




Please make your question self-contained instead of linking to another site.
– amd
Nov 26 at 23:54












That would be so lòng. Please check my edited link. Thanks
– Willy
Nov 27 at 13:08




That would be so lòng. Please check my edited link. Thanks
– Willy
Nov 27 at 13:08












Part of the function of this site is as an archive. External links go stale, so some day no one will have any idea what your question actually was when they come to this post. Aside from that, you’re asking others to take their time to help you. It’s only fair that you take some time of your own to formulate your question properly.
– amd
Nov 27 at 20:04




Part of the function of this site is as an archive. External links go stale, so some day no one will have any idea what your question actually was when they come to this post. Aside from that, you’re asking others to take their time to help you. It’s only fair that you take some time of your own to formulate your question properly.
– amd
Nov 27 at 20:04












Thanks @ and. Lesson learned. I edited my question.
– Willy
Nov 28 at 5:33




Thanks @ and. Lesson learned. I edited my question.
– Willy
Nov 28 at 5:33












The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $xne y$.
– amd
Nov 28 at 6:14






The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $xne y$.
– amd
Nov 28 at 6:14












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My first solution was wrong. Because unit vector isn't perpendicular to $v_1$, $v_2$.

My second solution: Dot product
$v_1$ . $v_2$ = |$v_1$| . |$v_2$| . cos 60 = $sqrt{6a^2 + 4}$/ 2 (1).
$v_1$ . $v_2$ = a.1 + (a+1).1 = 2a +1 (2)

(1) = (2) => a = -(8/5).

8 and 5 are coprime positive integers.

p + q = 13






share|cite|improve this answer





















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    My first solution was wrong. Because unit vector isn't perpendicular to $v_1$, $v_2$.

    My second solution: Dot product
    $v_1$ . $v_2$ = |$v_1$| . |$v_2$| . cos 60 = $sqrt{6a^2 + 4}$/ 2 (1).
    $v_1$ . $v_2$ = a.1 + (a+1).1 = 2a +1 (2)

    (1) = (2) => a = -(8/5).

    8 and 5 are coprime positive integers.

    p + q = 13






    share|cite|improve this answer


























      0














      My first solution was wrong. Because unit vector isn't perpendicular to $v_1$, $v_2$.

      My second solution: Dot product
      $v_1$ . $v_2$ = |$v_1$| . |$v_2$| . cos 60 = $sqrt{6a^2 + 4}$/ 2 (1).
      $v_1$ . $v_2$ = a.1 + (a+1).1 = 2a +1 (2)

      (1) = (2) => a = -(8/5).

      8 and 5 are coprime positive integers.

      p + q = 13






      share|cite|improve this answer
























        0












        0








        0






        My first solution was wrong. Because unit vector isn't perpendicular to $v_1$, $v_2$.

        My second solution: Dot product
        $v_1$ . $v_2$ = |$v_1$| . |$v_2$| . cos 60 = $sqrt{6a^2 + 4}$/ 2 (1).
        $v_1$ . $v_2$ = a.1 + (a+1).1 = 2a +1 (2)

        (1) = (2) => a = -(8/5).

        8 and 5 are coprime positive integers.

        p + q = 13






        share|cite|improve this answer












        My first solution was wrong. Because unit vector isn't perpendicular to $v_1$, $v_2$.

        My second solution: Dot product
        $v_1$ . $v_2$ = |$v_1$| . |$v_2$| . cos 60 = $sqrt{6a^2 + 4}$/ 2 (1).
        $v_1$ . $v_2$ = a.1 + (a+1).1 = 2a +1 (2)

        (1) = (2) => a = -(8/5).

        8 and 5 are coprime positive integers.

        p + q = 13







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 8:32









        Willy

        414




        414






























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