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) : $\left(1/\sqrt3; 1/\sqrt3; 1/\sqrt3\right)$

  1. Cross-product: $$ \begin{align} v_3 = v_1 \times v_2 &= \left(\sqrt{a^2+(a+1)^2+(a-1)^2} \cdot \sqrt2 \cdot \sin{60°}\right) \hat a \\ &= \left(\frac12\sqrt{6a^2+4}; \frac12\sqrt{6a^2+4}; \frac12\sqrt{6a^2+4}\right). \tag 1 \end{align}$$

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

From (1) and (2) => $$ \begin{align} &\; \frac12\sqrt{6a^2+4} = 1-a \\ &\iff\; \sqrt{6a^2+4} = 2-2a \\ &\iff\; a = -4 \text{ (checked)}. \end{align}$$

But $a = -p/q$ (coprime positive integers). So my solution is wrong.
Anyone knows how to solve it? Please give hint or instructions! Thanks!

amd
  • 53,693
Willy
  • 85
  • Please make your question self-contained instead of linking to another site. – amd Nov 26 '18 at 23:54
  • That would be so lòng. Please check my edited link. Thanks – Willy Nov 27 '18 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 '18 at 20:04
  • Thanks @ and. Lesson learned. I edited my question. – Willy Nov 28 '18 at 05:33
  • The two lines don’t actually intersect. If $z=1$, then we must have $x=1$ and $y=2$, but then $x\ne y$. – amd Nov 28 '18 at 06:14
  • Why did you introduce an extraneous unit vector? Why choose that particular direction for it? Leaving aside that the lines don’t intersect, you should be able to solve this by computing the dot product of the two direction vectors. – amd Nov 28 '18 at 06:19
  • How did you decide that $v_1\times v_2$ is a multiple of $\hat a$? The latter isn’t perpendicular to $v_2$, so it can’t possibly be the correct direction for this cross product. – amd Nov 28 '18 at 06:36
  • This site uses MathJax for formatting mathematical expressions. I’ve taken the liberty of editing your question to add this formatting, but please do this yourself in the future. You can find a quick reference and tutorial here. – amd Nov 28 '18 at 06:37
  • @amd. I can't thank enough for your kindness and enthusiasm. – Willy Nov 28 '18 at 08:35

1 Answers1

1

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

Willy
  • 85