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) : $\left(1/\sqrt3; 1/\sqrt3; 1/\sqrt3\right)$
Cross-product: $$ \begin{align} v_3 = v_1 \times v_2 &= \left(\sqrt{a^2+(a+1)^2+(a-1)^2} \cdot \sqrt2 \cdot \sin{60°}\right) \hat a \\ &= \left(\frac12\sqrt{6a^2+4}; \frac12\sqrt{6a^2+4}; \frac12\sqrt{6a^2+4}\right). \tag 1 \end{align}$$
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!
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