This one's probably easy, but I'm dreadfully stuck and can't seem to figure out a decent method.
I have the following lines:
$$a: \vec{x}(\lambda)= \left( \begin{array}{ccc} 4 \\ -2 \\ -2 \end{array} \right) + \lambda\left( \begin{array}{ccc} 1 \\ -1 \\ -1 \end{array} \right) $$ $$b: \vec{x}(\mu)= \left( \begin{array}{ccc} -1 \\ 1 \\ -3 \end{array} \right) + \mu\left( \begin{array}{ccc} 1 \\ 0 \\ 2 \end{array} \right) $$ $$c: \vec{x}(\nu)= \left( \begin{array}{ccc} 1 \\ 0 \\ 5 \end{array} \right) + \nu\left( \begin{array}{ccc} 0 \\ -2 \\ 1 \end{array} \right) $$ $$d: \vec{x}(\tau)= \left( \begin{array}{ccc} 3 \\ -2 \\ 0 \end{array} \right) + \tau\left( \begin{array}{ccc} -1 \\ 1 \\ 1 \end{array} \right) $$ I have to find the line that intersects all four of these lines. How do I go about doing this?
Cheers!
Note that $a$ and $d$ are parallel, so whatever line it is, it needs to lie in the plane containing $a$ and $d$. Considering it also needs to intersect $b$ and $c$, try figuring out which two points those two lines intersect the $ad$-plane.
Edit: Here's a full answer.
$ad$-plane: The $ad$-plane's normal vector is orthogonal to $(-1, 1, 1)$ as well as $\vec{a}(0) - \vec{d}(0) = (1, 0, -2)$. We therefore have a normal vector given by $$ (-1, 1, 1)\times (1, 0, -2) \\ = (1\cdot(-2) - 1\cdot 0, 1\cdot 1 - (-1)\cdot (-2), (-1)\cdot 0 - 1\cdot 1)\\ = (-2, -1, -1) $$ I elect to choose the negative of this vector, for estethic reasons.
Inserting $\vec{d}(0)$ into the general equation for a plane, we have: $$ 2\cdot 3 + 1\cdot (-2) + 1 \cdot 0 = 4 $$ and therefore the $ad$-plane is given by $2x + y + z = 4$.
$b$-intersection: The $\mu$ for the point where the $b$-line intersects the $ad$-plane is given by $$ 2(-1 + \mu) + 1 -3 + 2\mu = 4\\ 4\mu = 8\\ \mu = 2 $$ so the intersection point is $B = \vec{b}(2) = (1, 1, 1)$.
$c$-intersection: The $\nu$ for the point where the $c$-line intersects the $ad$-plane is given by $$ 2\cdot 1 -2\nu + 5 + \nu = 4\\ -\nu = -3\\ \nu = 3 $$ so the intersection point is $C = \vec{c}(3) = (1, -6, 8)$.
The line: We need the line $\vec{l}(\gamma)$ that goes from $B$ to $C$. It is given by $$ \vec{l}'(\gamma) = B + \gamma(C - B)\\ = (1, 1, 1) + \gamma(0, -7, 7) $$ which I would like to rewrite to: $$ \vec{l}(\gamma) = (1, 1, 1) + \gamma(0, -1, 1) $$ For reference, the four intersection points are: